UPSC Maths Optional · Paper II · Full Proofs + PYQs 2000–2026
Every bounded sequence in \(\mathbb{R}\) has a convergent subsequence.
Step 1 Let \((x_n)\) be bounded: \(a\le x_n\le b\) for all \(n\). Bisect \([a,b]\) into \([a,c]\) and \([c,b]\) where \(c=(a+b)/2\).
Step 2 At least one half contains infinitely many terms of the sequence — call it \([a_1,b_1]\).
Step 3 Repeat: bisect \([a_1,b_1]\), choose the half with infinitely many terms: \([a_2,b_2]\). Continue to get nested intervals \([a_k,b_k]\) with \(|a_k-b_k|=(b-a)/2^k\to0\).
Step 4 By Cantor's Intersection Theorem (completeness of \(\mathbb{R}\)): \(\exists\,L\in\bigcap_k[a_k,b_k]\). Each \([a_k,b_k]\) contains infinitely many terms, so pick \(x_{n_k}\in[a_k,b_k]\) with \(n_1
Bounded monotone → BW gives convergent subsequence → monotone + Cauchy-like argument → whole sequence converges.
Convergent ⟹ Cauchy: If \(x_n\to L\): \(|x_m-x_n|\le|x_m-L|+|L-x_n|<\epsilon/2+\epsilon/2=\epsilon\) for \(m,n>N\). \(\blacksquare\)
Cauchy ⟹ Convergent: Step 1 Cauchy sequences are bounded (take \(\epsilon=1\): all terms eventually within 1 of \(x_N\)).
Step 2 By BW, Cauchy has a convergent subsequence \(x_{n_k}\to L\).
Step 3 For \(\epsilon>0\): choose \(N_1\) s.t. \(|x_m-x_n|<\epsilon/2\) for \(m,n>N_1\), and \(N_2\) s.t. \(|x_{n_k}-L|<\epsilon/2\) for \(k>N_2\). Then for \(n>N_1\) and \(k>\max(N_1,N_2)\): \(|x_n-L|\le|x_n-x_{n_k}|+|x_{n_k}-L|<\epsilon\). \(\blacksquare\)
Every bounded monotone sequence of real numbers converges.
Say \((x_n)\) is bounded increasing. Let \(L=\sup\{x_n\}\) (exists by completeness). For \(\epsilon>0\): by definition of sup, \(\exists N\) s.t. \(x_N>L-\epsilon\). Since \((x_n)\) is increasing: for all \(n>N\), \(L-\epsilon
Let \(f:[1,\infty)\to\mathbb{R}\) be positive, continuous, and decreasing. Then \(\sum_{n=1}^\infty f(n)\) converges iff \(\int_1^\infty f(x)\,dx<\infty\).
Since \(f\) is decreasing: on \([n,n+1]\), \(f(n+1)\le f(x)\le f(n)\). Integrate over \([n,n+1]\): \(f(n+1)\le\int_n^{n+1}f\le f(n)\).
Sum from \(n=1\) to \(N\): \(\sum_{n=2}^{N+1}f(n)\le\int_1^{N+1}f\le\sum_{n=1}^N f(n)\).
If integral converges: partial sums \(\sum f(n)\le f(1)+\int_1^\infty f<\infty\). Bounded increasing → converges.
If series converges: \(\int_1^{N+1}f\le\sum_{n=1}^N f(n)\le\sum_{n=1}^\infty f(n)<\infty\). Integral converges. \(\blacksquare\)
For \(\sum a_n>0\): if \(n\!\left(\dfrac{a_n}{a_{n+1}}-1\right)\to L\), then series converges if \(L>1\), diverges if \(L<1\).
If \(L>1\): choose \(1
\(R=\dfrac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}\).
Let \(L=\limsup|a_n|^{1/n}\). For the power series \(\sum a_n x^n\), apply Root test: \(\limsup|a_nx^n|^{1/n}=|x|L\). If \(|x|L<1\) (i.e. \(|x|<1/L=R\)): series converges absolutely. If \(|x|L>1\): terms \(\not\to0\), series diverges. So radius is \(R=1/L\). \(\blacksquare\)
If \(f:[a,b]\to\mathbb{R}\) is continuous and \(f(a)<\gamma
Step 1 Let \(a_1=a,b_1=b\). Set \(c_1=(a_1+b_1)/2\). If \(f(c_1)=\gamma\): done. If \(f(c_1)<\gamma\): set \(a_2=c_1,b_2=b_1\). If \(f(c_1)>\gamma\): set \(a_2=a_1,b_2=c_1\).
Step 2 Continue: get nested intervals \([a_n,b_n]\) with \(f(a_n)<\gamma Step 3 By completeness: \(a_n\to c\) and \(b_n\to c\). By continuity: \(f(a_n)\to f(c)\) and \(f(b_n)\to f(c)\). Since \(f(a_n)<\gamma\le f(b_n)\): taking limits, \(f(c)=\gamma\). \(\blacksquare\)
Every continuous function on a closed bounded interval is uniformly continuous.
Suppose not uniformly continuous: \(\exists\epsilon_0>0\) and sequences \(x_n,y_n\in[a,b]\) with \(|x_n-y_n|<1/n\) but \(|f(x_n)-f(y_n)|\ge\epsilon_0\).
By BW: \((x_n)\) has a subsequence \(x_{n_k}\to c\in[a,b]\). Then \(y_{n_k}\to c\) too (since \(|x_{n_k}-y_{n_k}|<1/n_k\to0\)). By continuity: \(f(x_{n_k})\to f(c)\) and \(f(y_{n_k})\to f(c)\). So \(|f(x_{n_k})-f(y_{n_k})|\to0\). Contradicts \(\ge\epsilon_0\). \(\blacksquare\)
If \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(\exists c\in(a,b)\) s.t. \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\).
Define \(g(x)=f(x)-\dfrac{f(b)-f(a)}{b-a}(x-a)\). Then \(g(a)=f(a)\) and \(g(b)=f(b)-\dfrac{f(b)-f(a)}{b-a}(b-a)=f(a)\). So \(g(a)=g(b)\). Since \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\), by Rolle's theorem: \(\exists c\) with \(g'(c)=0\). \(g'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a}=0\). \(\blacksquare\)
\(f(x)=\sum_{k=0}^n\dfrac{f^{(k)}(a)}{k!}(x-a)^k+\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\) for some \(\xi\) between \(a\) and \(x\).
Define \(R_n(x)=f(x)-\sum_{k=0}^n\dfrac{f^{(k)}(a)}{k!}(x-a)^k\) and \(g(t)=(x-t)^{n+1}\). Apply Cauchy MVT to \(F(t)=f(x)-\sum_{k=0}^n\dfrac{f^{(k)}(t)}{k!}(x-t)^k\) and \(G(t)=(x-t)^{n+1}\) on \([a,x]\).
\(F(x)=0\), \(G(x)=0\). Cauchy MVT: \(\dfrac{F(a)-F(x)}{G(a)-G(x)}=\dfrac{F'(\xi)}{G'(\xi)}\). Compute \(F'(t)=-\dfrac{f^{(n+1)}(t)}{n!}(x-t)^n\), \(G'(t)=-(n+1)(x-t)^n\). So \(\dfrac{F(a)}{G(a)}=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}\), giving \(R_n=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\). \(\blacksquare\)
Let \(f\) be continuous on \([a,b]\). By Cantor's theorem, \(f\) is uniformly continuous: \(\forall\epsilon>0,\exists\delta>0\) s.t. \(|x-y|<\delta\Rightarrow|f(x)-f(y)|<\epsilon/(b-a)\).
Choose a partition \(P=\{a=x_0<\cdots
If \(f\) is Riemann integrable on \([a,b]\) and \(F(x)=\int_a^x f(t)\,dt\), then \(F\) is continuous. If \(f\) is also continuous at \(x_0\), then \(F'(x_0)=f(x_0)\).
Continuity: \(|F(x)-F(y)|\le\int_{\min}^{\max}|f|\le M|x-y|\). So \(F\) is Lipschitz, hence continuous.
Differentiability at \(x_0\): \(\dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0)=\dfrac{1}{x-x_0}\int_{x_0}^x[f(t)-f(x_0)]\,dt\).
For \(\epsilon>0\): by continuity of \(f\) at \(x_0\), \(\exists\delta\) s.t. \(|t-x_0|<\delta\Rightarrow|f(t)-f(x_0)|<\epsilon\). For \(|x-x_0|<\delta\): \(\left|\dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0)\right|\le\epsilon\). So \(F'(x_0)=f(x_0)\). \(\blacksquare\)
Let \(f\) be monotone increasing on \([a,b]\). For uniform partition with \(n\) equal subintervals: \(M_i=f(x_i)\), \(m_i=f(x_{i-1})\).
\(U-L=\sum(f(x_i)-f(x_{i-1}))\cdot\dfrac{b-a}{n}=\dfrac{b-a}{n}[f(b)-f(a)]\to0\). By Darboux: \(f\) is Riemann integrable. \(\blacksquare\)
If \(|f_n(x)|\le M_n\) for all \(x\in S\) and \(\sum M_n<\infty\), then \(\sum f_n\) converges uniformly on \(S\).
Since \(\sum M_n<\infty\): the tails \(\sum_{n=N+1}^\infty M_n\to0\) as \(N\to\infty\). For any \(x\in S\):
\(\left|\sum_{n=N+1}^\infty f_n(x)\right|\le\sum_{n=N+1}^\infty|f_n(x)|\le\sum_{n=N+1}^\infty M_n\to0\) uniformly (bound is independent of \(x\)). So the series converges uniformly. \(\blacksquare\)
If \(f_n\to f\) uniformly on \([a,b]\) and each \(f_n\) is continuous, then \(f\) is continuous.
Fix \(x_0\). For \(\epsilon>0\): by uniform convergence, \(\exists N\) s.t. \(\sup|f_n-f|<\epsilon/3\) for all \(n>N\). Fix such \(n\). By continuity of \(f_n\): \(\exists\delta\) s.t. \(|x-x_0|<\delta\Rightarrow|f_n(x)-f_n(x_0)|<\epsilon/3\). Then:
\(|f(x)-f(x_0)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|<\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}=\epsilon\). \(\blacksquare\)
If \(f_n\to f\) uniformly on \([a,b]\) and each \(f_n\) is Riemann integrable, then \(\int_a^b f=\lim\int_a^b f_n\).
\(\left|\int_a^b f-\int_a^b f_n\right|\le\int_a^b|f-f_n|\le\sup|f-f_n|\cdot(b-a)\to0\) by uniform convergence. \(\blacksquare\)
Let \(f\in BV[a,b]\). Define \(p(x)=\dfrac{1}{2}[V_a^x(f)+f(x)-f(a)]\) and \(q(x)=\dfrac{1}{2}[V_a^x(f)-f(x)+f(a)]\). Then \(f(x)-f(a)=p(x)-q(x)\), and \(p,q\ge0\). Both are non-decreasing: \(p(y)-p(x)=\dfrac{1}{2}[V_x^y(f)+(f(y)-f(x))]\ge0\) since \(|f(y)-f(x)|\le V_x^y(f)\). Similarly for \(q\). So \(f=f(a)+p-q\) with \(p,q\) non-decreasing. \(\blacksquare\)
If \((X,d)\) is a complete metric space and \(T:X\to X\) is a contraction (\(d(Tx,Ty)\le kd(x,y)\), \(k<1\)), then \(T\) has a unique fixed point.
Step 1 Existence: Pick any \(x_0\). Define \(x_n=T^n x_0\). Then \(d(x_{n+1},x_n)\le k^n d(x_1,x_0)\). For \(m>n\): \(d(x_m,x_n)\le\dfrac{k^n}{1-k}d(x_1,x_0)\to0\). So \((x_n)\) is Cauchy.
Step 2 By completeness: \(x_n\to x^*\). By continuity of \(T\) (it's a contraction): \(Tx^*=T(\lim x_n)=\lim Tx_n=\lim x_{n+1}=x^*\).
Step 3 Uniqueness: If \(Tx^*=x^*\) and \(Ty^*=y^*\): \(d(x^*,y^*)=d(Tx^*,Ty^*)\le k\,d(x^*,y^*)\Rightarrow(1-k)d(x^*,y^*)\le0\Rightarrow d=0\). \(\blacksquare\)
A subset of \(\mathbb{R}^n\) is compact iff it is closed and bounded.
Compact ⟹ closed and bounded: Every compact metric space is complete (closed in any metric space) and totally bounded (bounded in \(\mathbb{R}\)).
Closed and bounded ⟹ compact: Let \(K\subset[a,b]\) be closed and bounded. Let \((x_n)\subset K\). By BW, has subsequence \(x_{n_k}\to L\). Since \(K\) is closed: \(L\in K\). So every sequence has a convergent subsequence in \(K\) → \(K\) is sequentially compact → compact (in \(\mathbb{R}\)). \(\blacksquare\)
Test for convergence: \(\displaystyle\sum_{n=1}^\infty\frac{n!}{n^n}\).
\(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{n!}=\dfrac{(n+1)n^n}{(n+1)^{n+1}}=\dfrac{n^n}{(n+1)^n}=\left(\dfrac{n}{n+1}\right)^n=\left(1-\dfrac{1}{n+1}\right)^n\to e^{-1}<1\). Series converges.
Show that \(f(x)=x\sin(\pi/x)\), \(f(0)=0\) is NOT of bounded variation on \([0,1]\).
Consider partition points \(x_k=2/(2k+1)\) for \(k=0,1,2,\ldots\) At these points \(\sin(\pi/x_k)=\sin((2k+1)\pi/2)=\pm1\) alternately, so \(f(x_k)=x_k\cdot(\pm1)=\pm x_k\). The variation over this partition: \(\sum|f(x_{k+1})-f(x_k)|\ge\sum(x_k+x_{k+1})\ge\sum 2/(2k+2)\sim\sum 1/k=\infty\). So \(V_0^1(f)=\infty\). Not BV. \(\blacksquare\)
Test \(\displaystyle\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\) for convergence.
Let \(a_n=\dfrac{(2n-1)!!}{(2n)!!}\). Ratio: \(\dfrac{a_n}{a_{n+1}}=\dfrac{2n+2}{2n+1}=1+\dfrac{1}{2n+1}\). Raabe: \(n\!\left(\dfrac{a_n}{a_{n+1}}-1\right)=\dfrac{n}{2n+1}\to\dfrac{1}{2}<1\). Series diverges.
Find the radius of convergence of \(\displaystyle\sum_{n=0}^\infty\frac{(2n)!}{(n!)^2}x^n\).
\(\dfrac{a_{n+1}}{a_n}=\dfrac{(2n+2)!}{((n+1)!)^2}\cdot\dfrac{(n!)^2}{(2n)!}\cdot|x|=\dfrac{(2n+2)(2n+1)}{(n+1)^2}|x|=\dfrac{4n+2}{n+1}|x|\to4|x|\). Converges if \(4|x|<1\Rightarrow R=1/4\).
Find \(\limsup\) and \(\liminf\) of \(a_n=(-1)^n\!\left(1+\dfrac{1}{n}\right)\).
For odd \(n\): \(a_n=-(1+1/n)\to-1\). For even \(n\): \(a_n=1+1/n\to1\). \(\limsup=1\), \(\liminf=-1\).
Discuss uniform convergence of \(f_n(x)=\dfrac{nx}{1+n^2x^2}\) on \([0,1]\).
Pointwise limit: \(f(x)=0\) for all \(x\in[0,1]\) (for \(x=0\): trivially 0; for \(x>0\): \(nx/(n^2x^2+1)\sim1/(nx)\to0\)).
\(\sup_{x\in[0,1]}|f_n(x)-0|=\sup\dfrac{nx}{1+n^2x^2}\). Maximum at \(x=1/n\): value \(=\dfrac{n/n}{1+1}=1/2\). So \(\sup=1/2\not\to0\). NOT uniformly convergent on \([0,1]\).
On \([\delta,1]\) for any \(\delta>0\): \(\sup\le\dfrac{1}{n\delta^2}\to0\). Uniformly convergent on \([\delta,1]\). \(\blacksquare\)
Test \(\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^p}\) for absolute and conditional convergence for various \(p>0\).
Absolute convergence: \(\sum 1/n^p\) converges iff \(p>1\). So series absolutely convergent for \(p>1\).
Conditional convergence: By Leibniz alternating test (terms decrease monotonically to 0 for \(p>0\)): series converges for all \(p>0\). For \(0
Find the radius of convergence and region of convergence of \(\displaystyle\sum_{n=1}^\infty\frac{x^n}{n^2}\).
By root test: \(\limsup|a_n|^{1/n}=\limsup(1/n^2)^{1/n}=1\). So \(R=1\). At \(x=1\): \(\sum 1/n^2\) converges (p-series, \(p=2>1\)). At \(x=-1\): \(\sum(-1)^n/n^2\) converges (absolutely). Converges on \([-1,1]\).
Show that \(\displaystyle\sum_{n=2}^\infty\frac{1}{n\ln n}\) diverges.
Apply integral test: \(f(x)=\dfrac{1}{x\ln x}\) is positive, continuous, decreasing on \([2,\infty)\). \(\int_2^\infty\dfrac{dx}{x\ln x}=[\ln(\ln x)]_2^\infty=\infty\). By integral test, the series diverges. \(\blacksquare\)
Show that the sequence \(x_n=1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^n}\) is a Cauchy sequence and find its limit.
For \(m>n\): \(|x_m-x_n|=\dfrac{1}{2^{n+1}}+\cdots+\dfrac{1}{2^m}<\dfrac{1}{2^{n+1}}\cdot\dfrac{1}{1-1/2}=\dfrac{1}{2^n}\to0\). Cauchy. Limit = \(\sum_{k=0}^\infty 1/2^k = 2\).
Test \(\displaystyle\sum_{n=1}^\infty\sin\!\left(\frac{1}{n}\right)\) for convergence.
Since \(\sin(1/n)\sim 1/n\) as \(n\to\infty\) (limit comparison: \(\dfrac{\sin(1/n)}{1/n}\to1>0\)), and \(\sum 1/n\) diverges, by limit comparison test: \(\sum\sin(1/n)\) diverges.
Test \(\displaystyle\sum_{n=1}^\infty\left(\frac{n}{n+1}\right)^{n^2}\) for convergence.
Root test: \(|a_n|^{1/n}=\left(\dfrac{n}{n+1}\right)^n=\left(1-\dfrac{1}{n+1}\right)^n\to e^{-1}<1\). Series converges absolutely.
Show that \(f(x)=x^2\) is uniformly continuous on \([0,2]\) but not on \([0,\infty)\).
On \([0,2]\): By Cantor's theorem (continuous on closed bounded interval ⟹ UC). Direct: \(|f(x)-f(y)|=|x+y||x-y|\le4|x-y|\). Given \(\epsilon\), take \(\delta=\epsilon/4\).
On \([0,\infty)\): Take \(x_n=n+1/n\), \(y_n=n\). \(|x_n-y_n|=1/n\to0\) but \(|x_n^2-y_n^2|=(2n+1/n)/n^2\to2\ne0\). Not UC. \(\blacksquare\)
Using MVT, show that \(\dfrac{b-a}{b}\le\ln\dfrac{b}{a}\le\dfrac{b-a}{a}\) for \(0
Apply MVT to \(f(x)=\ln x\) on \([a,b]\): \(\exists c\in(a,b)\) s.t. \(\ln b-\ln a=\dfrac{b-a}{c}\). Since \(a
Using Taylor's theorem, prove \(\cos x\ge1-\dfrac{x^2}{2}\) for all \(x\).
By Taylor with Lagrange remainder: \(\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots=1-\dfrac{x^2}{2!}+\dfrac{\cos\xi}{4!}x^4\) for some \(\xi\) between 0 and \(x\). Since \(\cos\xi\ge-1\) and \(x^4\ge0\): \(\cos x=1-\dfrac{x^2}{2}+\dfrac{\cos\xi}{4!}x^4\ge1-\dfrac{x^2}{2}-\dfrac{x^4}{24}\). But actually simpler: use \(\cos x=1-\dfrac{x^2}{2}+\dfrac{\cos\xi}{4!}x^4\ge1-\dfrac{x^2}{2}\) since \(\dfrac{\cos\xi}{4!}x^4\ge\dfrac{-1}{24}x^4\)... direct proof: integrate \(\cos t\le1\) from 0 to \(x\): \(\sin x\le x\). Integrate again: \(-\cos x\le x^2/2-1\), i.e. \(\cos x\ge1-x^2/2\). \(\blacksquare\)
Classify the discontinuity of \(f(x)=\dfrac{\sin x}{|x|}\) at \(x=0\).
\(\lim_{x\to0^+}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to0^-}\dfrac{\sin x}{-x}=-1\). Left and right limits both exist but are unequal (\(1\ne-1\)). So \(x=0\) is a jump discontinuity (discontinuity of first kind).
If \(f(x)=x^2\sin(1/x)\) for \(x\ne0\) and \(f(0)=0\), show \(f\) is differentiable at 0 but \(f'\) is not continuous at 0.
\(f'(0)=\lim_{h\to0}\dfrac{h^2\sin(1/h)}{h}=\lim h\sin(1/h)=0\). So \(f'(0)=0\). For \(x\ne0\): \(f'(x)=2x\sin(1/x)-\cos(1/x)\). As \(x\to0\): \(2x\sin(1/x)\to0\) but \(\cos(1/x)\) oscillates between -1 and 1. So \(\lim_{x\to0}f'(x)\) does not exist. \(f'\) is not continuous at 0. \(\blacksquare\)
Show that the equation \(x^5-5x+1=0\) has exactly one real root in \([0,1]\).
\(f(0)=1>0\), \(f(1)=-3<0\). By IVT: at least one root in \((0,1)\). \(f'(x)=5x^4-5=5(x^4-1)\le0\) on \([0,1]\) (since \(x^4\le1\)). So \(f\) is decreasing on \([0,1]\), hence strictly — at most one root. Exactly one root. \(\blacksquare\)
Prove that if \(|f(x)-f(y)|\le M|x-y|\) for all \(x,y\), then \(f\) is uniformly continuous.
Given \(\epsilon>0\): take \(\delta=\epsilon/M\). If \(|x-y|<\delta\): \(|f(x)-f(y)|\le M|x-y|
Show that the Dirichlet function \(f(x)=\begin{cases}1&x\in\mathbb{Q}\\0&x\notin\mathbb{Q}\end{cases}\) is NOT Riemann integrable on \([0,1]\).
For any partition \(P\) of \([0,1]\): each subinterval contains both rationals and irrationals (by density). So \(M_i=1\) and \(m_i=0\) for all \(i\). \(U(f,P)=1\) and \(L(f,P)=0\) for all partitions. Since \(U\ne L\), \(f\) is not Riemann integrable. \(\blacksquare\)
Evaluate \(\displaystyle\int_0^\infty\frac{\sin x}{x}\,dx=\frac{\pi}{2}\) using Laplace transform.
Let \(I(s)=\int_0^\infty e^{-sx}\dfrac{\sin x}{x}dx\). \(I'(s)=-\int_0^\infty e^{-sx}\sin x\,dx=-\dfrac{1}{s^2+1}\). Integrating: \(I(s)=-\arctan s+C\). As \(s\to\infty\): \(I(s)\to0\) and \(-\arctan s\to-\pi/2\Rightarrow C=\pi/2\). So \(I(s)=\pi/2-\arctan s\). At \(s=0\): \(I(0)=\pi/2\). \(\blacksquare\)
State and prove the Second Mean Value Theorem for integrals.
Statement: If \(f\) is monotone decreasing and \(g\) is Riemann integrable on \([a,b]\), then \(\exists c\in[a,b]\) s.t. \(\int_a^b fg=f(a)\int_a^c g+f(b)\int_c^b g\).
Proof: Let \(G(x)=\int_a^x g\). By Abel's summation (integration by parts analog for Stieltjes integral): \(\int_a^b fg=f(b)G(b)-\int_a^b G\,df\). Since \(f\) is monotone (hence BV), apply MVT to Stieltjes integral: \(\int_a^b G\,df=G(c)[f(b)-f(a)]\) for some \(c\in[a,b]\). Substituting gives the result. \(\blacksquare\)
If \(F(x)=\int_x^{x^2}\frac{\sin t}{t}\,dt\), find \(F'(x)\).
By FTC + chain rule (Leibniz): \(F'(x)=\dfrac{\sin(x^2)}{x^2}\cdot 2x-\dfrac{\sin x}{x}\cdot 1=\dfrac{2\sin(x^2)}{x}-\dfrac{\sin x}{x}\).
Test \(\displaystyle\int_0^1\frac{dx}{\sqrt{x(1-x)}}\) for convergence and evaluate if convergent.
Singularities at \(x=0\) and \(x=1\). Near \(x=0\): integrand \(\sim x^{-1/2}\), and \(\int_0^{1/2}x^{-1/2}dx<\infty\). Near \(x=1\): \(\sim(1-x)^{-1/2}\), similarly convergent. So the integral converges. Evaluation: substitute \(x=\sin^2\theta\): \(\int_0^{\pi/2}\dfrac{2\sin\theta\cos\theta\,d\theta}{\sin\theta\cos\theta}=\int_0^{\pi/2}2\,d\theta=\pi\).
Prove that a monotone function on \([a,b]\) is Riemann integrable.
WLOG \(f\) is increasing. For uniform partition with \(n\) parts: \(U-L=\sum(M_i-m_i)h=\sum(f(x_i)-f(x_{i-1}))\cdot\dfrac{b-a}{n}=\dfrac{b-a}{n}[f(b)-f(a)]\to0\). By Darboux criterion: \(f\) is Riemann integrable. \(\blacksquare\)
Evaluate \(\displaystyle\int_0^1\frac{x^a-x^b}{\ln x}\,dx\) where \(a,b>0\).
Let \(I(a)=\int_0^1\dfrac{x^a-x^b}{\ln x}dx\). \(I'(a)=\int_0^1 x^a\,dx=\dfrac{1}{a+1}\). Integrating: \(I(a)=\ln(a+1)+C\). At \(a=b\): \(I(b)=0\Rightarrow C=-\ln(b+1)\). So \(I(a)=\ln\dfrac{a+1}{b+1}\).
Show that \(\displaystyle\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}\) converges uniformly on \(\mathbb{R}\).
\(|f_n(x)|=\left|\dfrac{\cos(nx)}{n^2}\right|\le\dfrac{1}{n^2}=M_n\) for all \(x\in\mathbb{R}\). Since \(\sum M_n=\sum 1/n^2=\pi^2/6<\infty\), by Weierstrass M-test the series converges uniformly. \(\blacksquare\)
If \(f_n(x)=nxe^{-nx^2}\), find the pointwise limit and determine if \(\int_0^1\lim f_n=\lim\int_0^1 f_n\).
Pointwise: for \(x>0\): \(f_n(x)\to0\); for \(x=0\): \(f_n(0)=0\). So \(f\equiv0\). \(\int_0^1 0\,dx=0\).
\(\int_0^1 nxe^{-nx^2}dx=\left[-\dfrac{e^{-nx^2}}{2}\right]_0^1=\dfrac{1}{2}(1-e^{-n})\to\dfrac{1}{2}\ne0\). So interchange fails — convergence is NOT uniform on \([0,1]\).
Determine whether \(f_n(x)=\dfrac{x^n}{1+x^n}\) converges uniformly on \([0,1]\).
For \(x\in[0,1)\): \(f_n(x)\to0\). For \(x=1\): \(f_n(1)=1/2\). So pointwise limit \(f(x)=0\) for \(x\in[0,1)\) and \(f(1)=1/2\). Since \(f\) is discontinuous at \(x=1\) but each \(f_n\) is continuous: convergence is NOT uniform on \([0,1]\) (by the uniform limit theorem: uniform limit of continuous functions is continuous). \(\blacksquare\)
Show \(\displaystyle\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{2n+1}=\arctan x\) converges uniformly on \([-1,1]\).
For \(|x|\le1\): \(\left|\dfrac{(-1)^n x^{2n+1}}{2n+1}\right|\le\dfrac{1}{2n+1}=M_n\) and \(\sum M_n=\sum 1/(2n+1)\) diverges... so M-test fails. Use Abel's theorem: the power series \(\sum(-1)^n x^{2n+1}/(2n+1)\) converges for \(|x|<1\) and at \(x=\pm1\) by alternating series test. By Abel's theorem, it converges uniformly on \([-1,1]\). \(\blacksquare\)
If \(f_n(x)=\dfrac{1}{n}\sin(nx)\), find the pointwise limit and show \(\lim\int_0^\pi f_n\ne\int_0^\pi\lim f_n\). Explain.
Pointwise: \(f_n(x)=\sin(nx)/n\to0\). \(\int_0^\pi 0\,dx=0\). But \(\int_0^\pi\dfrac{\sin(nx)}{n}dx=\dfrac{1}{n}\left[-\dfrac{\cos(nx)}{n}\right]_0^\pi=\dfrac{1-\cos(n\pi)}{n^2}\to0\). Actually here both give 0! The issue arises with \(g_n=n\sin(nx)/n^2\)... let me re-examine: actually \(\int_0^\pi f_n=\dfrac{2}{n^2}(n\) odd\()\to0\). So here limits agree. The convergence IS uniform here: \(\sup|f_n|=1/n\to0\). \(\blacksquare\)
Show that \(\mathbb{Q}\) with usual metric is not complete.
The sequence of rational approximations to \(\sqrt2\): \(x_1=1, x_{n+1}=\dfrac{x_n}{2}+\dfrac{1}{x_n}\) converges in \(\mathbb{R}\) to \(\sqrt2\notin\mathbb{Q}\). But \((x_n)\subset\mathbb{Q}\) is a Cauchy sequence (convergent in \(\mathbb{R}\) ⟹ Cauchy). It has no limit in \(\mathbb{Q}\). So \(\mathbb{Q}\) is not complete. \(\blacksquare\)
Show that the set \(S=(0,1)\) is open but not closed in \(\mathbb{R}\). Show \([0,1]\) is closed but not open.
Open: For any \(x\in(0,1)\): take \(r=\min(x,1-x)>0\). Then \(B(x,r)\subset(0,1)\). So \((0,1)\) is open.
Not closed: \((0,1)\) does not contain its limit point \(0\) (sequence \(1/n\to0\notin(0,1)\)).
\([0,1]\) closed: It equals its own closure (contains all its limit points).
\([0,1]\) not open: No open ball around \(0\) is contained in \([0,1]\) (every ball contains negative numbers). \(\blacksquare\)
Use the Banach fixed point theorem to prove \(\cos x=x\) has a unique solution in \([0,\pi/2]\).
Let \(T(x)=\cos x\) on \(I=[0,\pi/2]\). \(T:I\to I\) since \(\cos0=1,\cos(\pi/2)=0\). \(|T(x)-T(y)|=|\cos x-\cos y|\le|x-y|\cdot|\sin\xi|\le|x-y|\cdot1\). But we need strict contraction. On \([0,1]\): \(|\sin\xi|\le\sin1<1\). So \(T\) is a contraction on \([0,1]\) (which maps to itself: \(\cos([0,1])\subset[0,1]\)). By Banach's theorem: unique fixed point. \(\blacksquare\)
Show that every compact metric space is complete.
Let \((x_n)\) be a Cauchy sequence in compact \((X,d)\). By compactness: \((x_n)\) has a convergent subsequence \(x_{n_k}\to L\). Since \((x_n)\) is Cauchy: for \(\epsilon>0\), choose \(N\) s.t. \(d(x_m,x_n)<\epsilon/2\) for \(m,n>N\), and \(K\) s.t. \(d(x_{n_k},L)<\epsilon/2\) for \(k>K\). Then for \(n>N\) and \(k>\max(K,N)\): \(d(x_n,L)\le d(x_n,x_{n_k})+d(x_{n_k},L)<\epsilon\). So \(x_n\to L\in X\). \(\blacksquare\)
Prove that every monotone function on \([a,b]\) is of bounded variation.
WLOG \(f\) is increasing. For any partition \(P=\{a=x_0<\cdots \(\sum_{i=1}^n|f(x_i)-f(x_{i-1})|=\sum(f(x_i)-f(x_{i-1}))=f(b)-f(a)\). Since this is independent of the partition: \(V_a^b(f)=f(b)-f(a)<\infty\). So \(f\in BV[a,b]\). \(\blacksquare\)
Prove that \(\mathbb{Q}\) is dense in \(\mathbb{R}\), i.e., every non-empty open interval \((a,b)\) contains a rational number.
Since \(b-a>0\): by Archimedean property, \(\exists n\in\mathbb{N}\) with \(1/n
If \(f\in BV[a,b]\), write \(f=f_1-f_2\) where both \(f_1,f_2\) are non-decreasing.
Take \(f_1(x)=V_a^x(f)\) (total variation function) and \(f_2(x)=V_a^x(f)-f(x)+f(a)\). Both are non-decreasing and \(f_1-f_2=f-f(a)\). Adding \(f(a)\) as a constant gives \(f=V_a^x(f)-[V_a^x(f)-f(x)]=f_1-(f_1-f)\), which works since \(f_1-f\) is also increasing. \(\blacksquare\)
Prove that \(d(x,y)=\dfrac{|x-y|}{1+|x-y|}\) defines a metric on \(\mathbb{R}\).
M1 (positivity): \(d(x,y)\ge0\); \(d(x,y)=0\iff|x-y|=0\iff x=y\). M2 (symmetry): \(d(x,y)=d(y,x)\) since \(|x-y|=|y-x|\). M3 (triangle inequality): Use the fact that \(f(t)=t/(1+t)\) is increasing and sub-additive: if \(t\le u+v\) then \(f(t)\le f(u+v)\le f(u)+f(v)\) (last inequality: \(\frac{u+v}{1+u+v}\le\frac{u}{1+u}+\frac{v}{1+v}\) since LHS denominator \(\ge\) each summand's denominator). With \(t=|x-z|\), \(u=|x-y|\), \(v=|y-z|\) and \(|x-z|\le|x-y|+|y-z|\): \(d(x,z)\le d(x,y)+d(y,z)\). \(\blacksquare\)
State and prove the Cauchy-Schwarz inequality in the sequence space \(\ell^2\): for \((a_n),(b_n)\in\ell^2\), show \(\left|\sum a_n b_n\right|\le\left(\sum a_n^2\right)^{1/2}\left(\sum b_n^2\right)^{1/2}\).
For any \(\lambda\in\mathbb{R}\): \(0\le\sum(a_n+\lambda b_n)^2=\sum a_n^2+2\lambda\sum a_n b_n+\lambda^2\sum b_n^2\). This is a non-negative quadratic in \(\lambda\), so its discriminant is \(\le0\): \(4\left(\sum a_n b_n\right)^2-4\left(\sum a_n^2\right)\left(\sum b_n^2\right)\le0\). Hence \(\left(\sum a_n b_n\right)^2\le\left(\sum a_n^2\right)\left(\sum b_n^2\right)\). (All finite partial sums satisfy this; the infinite series case follows by taking limits since the series converge absolutely.) \(\blacksquare\)
Find the interior, closure, and boundary of \(\mathbb{Q}\) in \(\mathbb{R}\) with the usual metric.
Interior \(\mathbb{Q}^\circ=\emptyset\): Every open interval \((a,b)\) contains irrationals, so no open ball sits entirely inside \(\mathbb{Q}\). Closure \(\overline{\mathbb{Q}}=\mathbb{R}\): \(\mathbb{Q}\) is dense in \(\mathbb{R}\); every real number is a limit of rationals, so every point of \(\mathbb{R}\) is a limit point of \(\mathbb{Q}\). Boundary \(\partial\mathbb{Q}=\mathbb{R}\): \(\partial\mathbb{Q}=\overline{\mathbb{Q}}\setminus\mathbb{Q}^\circ=\mathbb{R}\setminus\emptyset=\mathbb{R}\). \(\blacksquare\)
Define interior, closure, and boundary of a set \(E\) in a metric space. Prove \(\partial E=\overline{E}\setminus E^\circ\).
\(E^\circ=\bigcup\{U:U\text{ open},U\subset E\}\); \(\overline{E}=\bigcap\{F:F\text{ closed},F\supset E\}\); \(\partial E=\overline{E}\cap\overline{E^c}\). A point \(x\in\partial E\) iff every open ball \(B(x,r)\) meets both \(E\) and \(E^c\), i.e., \(x\notin E^\circ\) and \(x\notin(E^c)^\circ=(E^\circ)^c\cap\cdots\). More directly: \(x\in\overline{E}\) means every ball meets \(E\); \(x\notin E^\circ\) means no ball is contained in \(E\). Together these say every ball meets \(E^c\), so \(x\in\overline{E^c}\). Hence \(\partial E=\overline{E}\cap\overline{E^c}=\overline{E}\setminus E^\circ\) (since \(x\in\overline{E}\) and \(x\notin E^\circ\) iff \(x\in\overline{E}\cap(E^\circ)^c\)). \(\blacksquare\)
Show that \(d_1(x,y)=|x_1-y_1|+|x_2-y_2|\) defines a metric on \(\mathbb{R}^2\). Describe the open ball \(B(0,1)\).
Metric axioms: Positivity and symmetry are clear from properties of \(|\cdot|\). Triangle inequality: \(d_1(x,z)=|x_1-z_1|+|x_2-z_2|\le(|x_1-y_1|+|y_1-z_1|)+(|x_2-y_2|+|y_2-z_2|)=d_1(x,y)+d_1(y,z)\). Open ball: \(B(0,1)=\{(x_1,x_2):|x_1|+|x_2|<1\}\), a diamond (square rotated 45°) with vertices at \((\pm1,0)\) and \((0,\pm1)\). \(\blacksquare\)
Show that the Cantor set \(\mathcal{C}\) is perfect, nowhere dense, and uncountable.
Perfect (closed, every point is a limit point): \(\mathcal{C}=\bigcap_{n=0}^\infty C_n\) is an intersection of closed sets, hence closed. Every point of \(\mathcal{C}\) is an endpoint of one of the removed intervals, and endpoints are never removed; endpoints accumulate at each point of \(\mathcal{C}\). Nowhere dense: \(\mathcal{C}\) contains no interval (each stage removes middle thirds; any open interval eventually falls in a removed part). So \(\mathcal{C}^\circ=\emptyset\) and \(\overline{\mathcal{C}}^\circ=\emptyset\). Uncountable: The ternary representation gives a bijection between \(\mathcal{C}\) and \(\{0,2\}^{\mathbb{N}}\cong\{0,1\}^{\mathbb{N}}\), which has the cardinality of \(\mathbb{R}\). \(\blacksquare\)
State and prove the Heine-Borel theorem: A subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded.
(\(\Rightarrow\)) Compact \(\Rightarrow\) closed and bounded: Compact subsets of Hausdorff spaces are closed. For boundedness: \(\{B(0,n):n\in\mathbb{N}\}\) is an open cover; by compactness a finite subcover exists, bounding the set. (\(\Leftarrow\)) Closed and bounded \(\Rightarrow\) compact: A closed and bounded set is contained in a closed box \([-M,M]^n\). By bisection (or Tychonoff for cubes), the box is compact. A closed subset of a compact set is compact. \(\blacksquare\)
State the Bolzano-Weierstrass theorem. Use it to show every bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence.
Statement: Every bounded infinite set in \(\mathbb{R}^n\) has a limit point. For sequences: Let \((x^{(k)})\) be bounded in \(\mathbb{R}^n\). By Heine-Borel, the closure of \(\{x^{(k)}\}\) is compact. Extract a subsequence: apply BWT component-wise using the 1D version (bisection argument). For 1D: \((x_n)\subset[a,b]\); bisect repeatedly, always choosing the half containing infinitely many terms — the resulting nested intervals shrink to a point \(L\), and the chosen indices form a convergent subsequence \(x_{n_k}\to L\). Repeat for each coordinate of \(\mathbb{R}^n\). \(\blacksquare\)
Prove that in a metric space, a set is compact if and only if it is sequentially compact.
(Compact \(\Rightarrow\) Sequentially compact): Let \((x_n)\subset K\) compact. If the range is finite, some value repeats infinitely — a constant convergent subsequence. If infinite, by compactness it has a limit point \(L\in K\); build a subsequence converging to \(L\) by picking \(x_{n_k}\in B(L,1/k)\). (Sequentially compact \(\Rightarrow\) Compact): Show totally bounded (every \(\epsilon\)-net is finite) and complete. Given any open cover, use total boundedness to extract a Lebesgue number argument, yielding a finite subcover. \(\blacksquare\)
Prove that the connected subsets of \(\mathbb{R}\) are precisely the intervals (including rays and the whole line).
(Intervals are connected): Suppose \(I\) is an interval and \(I=A\cup B\) with \(A,B\) non-empty, open in \(I\), disjoint. Pick \(a\in A\), \(b\in B\), say \(a
Prove that every path-connected metric space is connected.
Suppose \(X\) is path-connected but not connected. Then \(X=U\cup V\) with \(U,V\) non-empty, open, disjoint. Pick \(u\in U\), \(v\in V\). By path-connectedness \(\exists\) continuous \(\gamma:[0,1]\to X\) with \(\gamma(0)=u\), \(\gamma(1)=v\). Then \([0,1]=\gamma^{-1}(U)\cup\gamma^{-1}(V)\), a disconnection of \([0,1]\) into two non-empty open sets — contradiction (\([0,1]\) is connected). \(\blacksquare\)
Define the completion of a metric space. State that every metric space has a unique (up to isometry) completion.
Definition: A complete metric space \((\hat{X},\hat{d})\) is a completion of \((X,d)\) if there exists an isometric embedding \(\iota:X\to\hat{X}\) with \(\iota(X)\) dense in \(\hat{X}\). Construction: Take equivalence classes of Cauchy sequences in \(X\): \((x_n)\sim(y_n)\) iff \(d(x_n,y_n)\to0\). Define \(\hat{d}([(x_n)],[(y_n)])=\lim d(x_n,y_n)\) (this limit exists). Embed \(X\) via constant sequences. The resulting \(\hat{X}\) is complete and \(\iota(X)\) is dense. Uniqueness: Any two completions are isometric via the unique extension of the identity on \(X\). \(\blacksquare\)
Prove the Banach fixed-point theorem: if \(T:(X,d)\to(X,d)\) is a contraction on a complete metric space, then \(T\) has a unique fixed point.
Let \(k\in[0,1)\) be the contraction constant. Fix any \(x_0\in X\); define \(x_{n+1}=T(x_n)\). Existence (Cauchy sequence): \(d(x_{n+1},x_n)\le k^n d(x_1,x_0)\). For \(m>n\): \(d(x_m,x_n)\le\dfrac{k^n}{1-k}d(x_1,x_0)\to0\). So \((x_n)\) is Cauchy; by completeness \(x_n\to L\). By continuity of \(T\): \(T(L)=T(\lim x_n)=\lim T(x_n)=\lim x_{n+1}=L\). Uniqueness: If \(L'=T(L')\) then \(d(L,L')=d(T(L),T(L'))\le k\,d(L,L')\Rightarrow d(L,L')=0\Rightarrow L=L'\). \(\blacksquare\)
State the Baire Category Theorem. Give an application.
Statement: In a complete metric space, the intersection of countably many dense open sets is dense. Equivalently, a complete metric space is of second category (cannot be written as a countable union of nowhere dense sets). Application: \(\mathbb{R}\) is of second category. Hence \(\mathbb{R}\) cannot be written as a countable union of nowhere dense sets. In particular, \(\mathbb{Q}=\bigcup_{q\in\mathbb{Q}}\{q\)\} cannot be a countable union of dense sets: each singleton \(\{q\}\) is nowhere dense, but their union is \(\mathbb{Q}\), confirming \(\mathbb{Q}\) is first category (meager) — but \(\mathbb{R}\setminus\mathbb{Q}\) is not. Also: a complete metric space with no isolated points is uncountable (else it would be meager). \(\blacksquare\)
Define nowhere dense sets, first category, and second category. Show that \(\mathbb{Q}\) is first category in \(\mathbb{R}\).
Nowhere dense: \(E\) is nowhere dense if \(\overline{E}\) has empty interior, i.e., \((\overline{E})^\circ=\emptyset\). First category (meager): \(E\) is a countable union of nowhere dense sets. Second category: Not first category. \(\mathbb{Q}\) is first category: \(\mathbb{Q}=\bigcup_{n=1}^\infty\{q_n\}\) (enumerated). Each singleton \(\{q_n\}\) is nowhere dense (\(\overline{\{q_n\}}=\{q_n\}\), which has empty interior). So \(\mathbb{Q}\) is a countable union of nowhere dense sets. By Baire, \(\mathbb{R}\) is second category, confirming \(\mathbb{R}\setminus\mathbb{Q}\) is not meager. \(\blacksquare\)
Prove: \(f:(X,d_X)\to(Y,d_Y)\) is continuous iff \(f^{-1}(V)\) is open in \(X\) for every open \(V\subseteq Y\).
(\(\Rightarrow\)): Let \(V\subset Y\) open, \(x\in f^{-1}(V)\). Then \(f(x)\in V\); since \(V\) open, \(\exists\epsilon>0\) with \(B(f(x),\epsilon)\subset V\). By continuity: \(\exists\delta>0\) s.t. \(d_X(x,y)<\delta\Rightarrow d_Y(f(x),f(y))<\epsilon\), i.e., \(B(x,\delta)\subset f^{-1}(V)\). So \(f^{-1}(V)\) is open. (\(\Leftarrow\)): Fix \(x\in X\), \(\epsilon>0\). \(V=B(f(x),\epsilon)\) is open, so \(f^{-1}(V)\) is open and contains \(x\). Thus \(\exists\delta>0\) with \(B(x,\delta)\subset f^{-1}(V)\), giving \(d_X(x,y)<\delta\Rightarrow f(y)\in V\Rightarrow d_Y(f(x),f(y))<\epsilon\). \(\blacksquare\)
Prove: \(f:(X,d_X)\to(Y,d_Y)\) is uniformly continuous iff it maps Cauchy sequences to Cauchy sequences.
(\(\Rightarrow\)): Let \((x_n)\) be Cauchy and \(\epsilon>0\). By uniform continuity: \(\exists\delta>0\) s.t. \(d_X(x,y)<\delta\Rightarrow d_Y(f(x),f(y))<\epsilon\). Since \((x_n)\) is Cauchy: \(\exists N\) s.t. \(m,n>N\Rightarrow d_X(x_m,x_n)<\delta\Rightarrow d_Y(f(x_m),f(x_n))<\epsilon\). So \((f(x_n))\) is Cauchy. (\(\Leftarrow\)): Suppose \(f\) is not uniformly continuous. Then \(\exists\epsilon_0>0\) and sequences \((x_n),(y_n)\) with \(d_X(x_n,y_n)<1/n\) but \(d_Y(f(x_n),f(y_n))\ge\epsilon_0\). Interleave: \(z_1=x_1,z_2=y_1,\ldots\) — the mixed sequence is Cauchy but \((f(z_n))\) is not. Contradiction. \(\blacksquare\)
Define an isometry between metric spaces. Show that isometries are injective and continuous.
Definition: \(f:(X,d_X)\to(Y,d_Y)\) is an isometry if \(d_Y(f(x),f(x'))=d_X(x,x')\) for all \(x,x'\in X\). Injective: If \(f(x)=f(x')\) then \(d_X(x,x')=d_Y(f(x),f(x'))=0\Rightarrow x=x'\). Continuous (uniformly): \(d_Y(f(x),f(x'))=d_X(x,x')<\epsilon\) whenever \(d_X(x,x')<\delta=\epsilon\). So \(f\) is an isometry with modulus of continuity \(\omega(\delta)=\delta\), i.e., 1-Lipschitz in both directions. \(\blacksquare\)
Define totally bounded. Prove: every compact metric space is totally bounded.
Definition: \((X,d)\) is totally bounded if for every \(\epsilon>0\) there exist finitely many points \(x_1,\ldots,x_n\in X\) such that \(X=\bigcup_{i=1}^n B(x_i,\epsilon)\). Proof: Let \(K\) be compact and \(\epsilon>0\). The collection \(\{B(x,\epsilon):x\in K\}\) is an open cover of \(K\). By compactness, finitely many suffice: \(K\subset\bigcup_{i=1}^n B(x_i,\epsilon)\). So \(K\) is totally bounded. \(\blacksquare\)
State the Arzelà-Ascoli theorem. Define equicontinuity.
Equicontinuous: A family \(\mathcal{F}\) of functions on \([a,b]\) is equicontinuous if for every \(\epsilon>0\), \(\exists\delta>0\) (independent of \(f\in\mathcal{F}\) and \(x\)) such that \(|x-y|<\delta\Rightarrow|f(x)-f(y)|<\epsilon\) for all \(f\in\mathcal{F}\). Arzelà-Ascoli: A subset \(\mathcal{F}\subset C([a,b])\) is relatively compact (i.e., every sequence has a uniformly convergent subsequence) if and only if \(\mathcal{F}\) is uniformly bounded and equicontinuous. \(\blacksquare\)
Suppose \((f_n)\) is a uniformly bounded, equicontinuous sequence on \([a,b]\). Show it has a uniformly convergent subsequence.
This is a consequence of Arzelà-Ascoli. Sketch: Let \(D=\{q_1,q_2,\ldots\}\) be a countable dense subset of \([a,b]\) (e.g., rationals). By uniform boundedness, \((f_n(q_1))\) is bounded — extract a convergent subsequence \((f_{n_k})\). From this, extract a further subsequence converging at \(q_2\), etc. By diagonalization, get \((f_{n_k})\) converging at every \(q_j\). By equicontinuity: for any \(x\in[a,b]\) and \(\epsilon>0\), choose \(q_j\) close to \(x\) and use equicontinuity to control \(|f_{n_k}(x)-f_{n_j}(x)|\). The convergence is uniform by compactness of \([a,b]\). \(\blacksquare\)
Describe how compactness in \(C([a,b])\) (with sup norm) differs from compactness in \(\mathbb{R}^n\). Why doesn't Heine-Borel apply directly?
In \(\mathbb{R}^n\), compact \(=\) closed and bounded (Heine-Borel). In infinite-dimensional spaces like \(C([a,b])\), closed and bounded does NOT imply compact. Example: \(f_n(x)=\sin(nx)\) is bounded (\(\|f_n\|_\infty=1\)) but has no uniformly convergent subsequence (for any \(n\ne m\), \(\|f_n-f_m\|_\infty=2\) at some point). The unit ball in \(C([a,b])\) is not compact. Compactness in \(C([a,b])\) requires the additional condition of equicontinuity, captured by Arzelà-Ascoli. \(\blacksquare\)
Prove: every bounded monotone sequence in \(\mathbb{R}\) converges.
WLOG \((x_n)\) is increasing and bounded above. Let \(L=\sup\{x_n\}\) (exists by completeness of \(\mathbb{R}\)). Given \(\epsilon>0\): \(L-\epsilon\) is not an upper bound, so \(\exists N\) with \(x_N>L-\epsilon\). For \(n\ge N\): \(L-\epsilon
Prove that \(x_n=\left(1+\dfrac{1}{n}\right)^n\) is bounded and increasing, hence convergent. Its limit is denoted \(e\).
Increasing: By AM-GM, \(\dfrac{1}{n+1}\sum_{k=0}^{n}\left(1+\dfrac{1}{n}\right)^{k/???}\)... use binomial: \(x_n=\sum_{k=0}^n\binom{n}{k}\dfrac{1}{n^k}\). Each term \(\binom{n}{k}/n^k=\dfrac{1}{k!}\cdot\dfrac{n(n-1)\cdots(n-k+1)}{n^k}\le\dfrac{1}{k!}\) and is increasing in \(n\). Adding more terms also increases the sum, so \(x_n
Prove that every Cauchy sequence in \(\mathbb{R}\) converges.
Step 1 (bounded): Take \(\epsilon=1\); \(\exists N\) s.t. \(n,m>N\Rightarrow|x_n-x_m|<1\). So \(|x_n|\le|x_N|+1\) for \(n>N\); together with the first \(N\) terms, \((x_n)\) is bounded. Step 2 (convergent subsequence): By Bolzano-Weierstrass, \(\exists x_{n_k}\to L\). Step 3 (whole sequence): Given \(\epsilon>0\): \(\exists N\) s.t. \(|x_m-x_n|<\epsilon/2\) for \(m,n>N\); and \(K\) s.t. \(|x_{n_k}-L|<\epsilon/2\) for \(k>K\). For \(n>N\) and \(k>\max(K,N)\): \(|x_n-L|\le|x_n-x_{n_k}|+|x_{n_k}-L|<\epsilon\). \(\blacksquare\)
Prove: \(x_n\to L\) iff every subsequence of \((x_n)\) has a further subsequence converging to \(L\).
(\(\Rightarrow\)): If \(x_n\to L\), every subsequence also converges to \(L\), so trivially true. (\(\Leftarrow\)): Suppose \(x_n\not\to L\). Then \(\exists\epsilon_0>0\) and infinitely many \(n_k\) with \(|x_{n_k}-L|\ge\epsilon_0\). The subsequence \((x_{n_k})\) has no further subsequence converging to \(L\) (all its terms are at distance \(\ge\epsilon_0\) from \(L\)). Contradiction. \(\blacksquare\)
Define \(\limsup\) and \(\liminf\) of a sequence. Compute them for \(x_n=(-1)^n\!\left(1+\dfrac{1}{n}\right)\).
Definitions: \(\limsup x_n=\inf_{n}\sup_{k\ge n}x_k=\lim_{n\to\infty}\sup_{k\ge n}x_k\); \(\liminf x_n=\sup_n\inf_{k\ge n}x_k\). Computation: For even \(n=2m\): \(x_{2m}=1+\tfrac{1}{2m}\to1^+\). For odd \(n=2m+1\): \(x_{2m+1}=-(1+\tfrac{1}{2m+1})\to-1^-\). The subsequential limits are \(1\) (even) and \(-1\) (odd), so \(\limsup x_n=1\) and \(\liminf x_n=-1\). Since \(\limsup\ne\liminf\), \((x_n)\) diverges. \(\blacksquare\)
Prove: if \(\sum a_n\) converges, then \(a_n\to0\). Show the converse is false.
Proof: Let \(S_n=\sum_{k=1}^n a_k\to S\). Then \(a_n=S_n-S_{n-1}\to S-S=0\). Converse false: The harmonic series \(\sum 1/n\): \(1/n\to0\) but \(\sum 1/n=\infty\) (diverges). Proof of divergence: \(H_{2^n}\ge1+n/2\to\infty\) by grouping. \(\blacksquare\)
Prove the Ratio Test: if \(\limsup|a_{n+1}/a_n|<1\), then \(\sum a_n\) converges absolutely.
Let \(L=\limsup|a_{n+1}/a_n|<1\). Choose \(r\in(L,1)\). Then \(\exists N\) s.t. \(|a_{n+1}/a_n|
Prove the Root Test: \(\sum a_n\) converges absolutely if \(\limsup|a_n|^{1/n}<1\), diverges if \(\limsup|a_n|^{1/n}>1\).
Let \(\alpha=\limsup|a_n|^{1/n}\). If \(\alpha<1\): Choose \(r\in(\alpha,1)\). Then \(|a_n|^{1/n}
State and prove the Direct Comparison Test for series with non-negative terms.
Statement: If \(0\le a_n\le b_n\) for all \(n\), then: (i) \(\sum b_n<\infty\Rightarrow\sum a_n<\infty\); (ii) \(\sum a_n=\infty\Rightarrow\sum b_n=\infty\). Proof (i): Partial sums \(A_N=\sum_{n=1}^N a_n\le\sum_{n=1}^N b_n\le\sum_{n=1}^\infty b_n<\infty\). So \((A_N)\) is increasing and bounded, hence convergent. (ii) follows by contrapositive. \(\blacksquare\)
State and prove the Integral Test: if \(f\) is positive, decreasing, and continuous on \([1,\infty)\), then \(\sum_{n=1}^\infty f(n)\) and \(\int_1^\infty f(x)\,dx\) converge or diverge together.
Since \(f\) is decreasing: \(f(n+1)\le f(x)\le f(n)\) for \(x\in[n,n+1]\). Integrating: \(f(n+1)\le\int_n^{n+1}f(x)\,dx\le f(n)\). Summing from \(n=1\) to \(N-1\): \(\sum_{n=2}^N f(n)\le\int_1^N f(x)\,dx\le\sum_{n=1}^{N-1}f(n)\). If \(\int_1^\infty f<\infty\): \(\sum f(n)\le f(1)+\int_1^\infty f<\infty\). If \(\sum f(n)<\infty\): \(\int_1^N f\le\sum_{n=1}^{N-1}f(n)\le\sum_{n=1}^\infty f(n)<\infty\). \(\blacksquare\)
Prove that \(\sum_{n=1}^\infty\dfrac{1}{n^p}\) converges if \(p>1\) and diverges if \(p\le1\).
Apply the Integral Test with \(f(x)=x^{-p}\). \(\int_1^\infty x^{-p}\,dx=\dfrac{x^{1-p}}{1-p}\Big|_1^\infty\) (for \(p\ne1\)): this equals \(\dfrac{1}{p-1}\) if \(p>1\) (convergent), and \(+\infty\) if \(p<1\) (divergent). For \(p=1\): \(\int_1^\infty\dfrac{dx}{x}=\infty\) (divergent). So \(\sum 1/n^p\) converges \(\iff p>1\). \(\blacksquare\)
State and prove the Alternating Series Test. Apply to \(\sum(-1)^{n+1}/n\).
Statement: If \((b_n)\) is positive, decreasing, and \(b_n\to0\), then \(\sum(-1)^{n+1}b_n\) converges. Proof: Even partial sums: \(S_{2n}=\sum_{k=1}^{2n}(-1)^{k+1}b_k\). Note \(S_{2n}=S_{2n-2}+b_{2n-1}-b_{2n}\ge S_{2n-2}\) (increasing). Also \(S_{2n}=b_1-(b_2-b_3)-\cdots-b_{2n}\le b_1\) (bounded). So \(S_{2n}\to L\). Odd partial sums: \(S_{2n+1}=S_{2n}+b_{2n+1}\to L+0=L\). Hence the series converges. Application: \(\sum(-1)^{n+1}/n=\ln2\) (conditionally convergent). \(\blacksquare\)
Define absolute and conditional convergence. Give an example of each and prove the relationship.
Absolute convergence: \(\sum a_n\) converges absolutely if \(\sum|a_n|<\infty\). Conditional: \(\sum a_n\) converges but \(\sum|a_n|=\infty\). Absolute \(\Rightarrow\) Convergent: \(|S_m-S_n|=\left|\sum_{k=n+1}^m a_k\right|\le\sum_{k=n+1}^m|a_k|\to0\), so \((S_n)\) is Cauchy. Examples: \(\sum 1/n^2\) converges absolutely; \(\sum(-1)^{n+1}/n\) converges conditionally (diverges absolutely by \(p\)-series). Converse fails: absolutely convergent is a strictly smaller class. \(\blacksquare\)
For \(\sum n^{-s}\) (Dirichlet series with \(a_n=1\)), find the abscissa of convergence and describe convergence behavior.
The Riemann zeta function \(\zeta(s)=\sum_{n=1}^\infty n^{-s}\) converges for \(s>1\) (real case) and diverges for \(s\le1\). More generally, a Dirichlet series \(\sum a_n n^{-s}\) has an abscissa of convergence \(\sigma_c\) such that the series converges for \(\text{Re}(s)>\sigma_c\) and diverges for \(\text{Re}(s)<\sigma_c\). For \(\zeta\): conditional convergence for \(s\in(0,1]\) does not hold (the series diverges for real \(s\le1\)); for complex \(s\) with \(\text{Re}(s)>1\) it converges absolutely. \(\blacksquare\)
State Abel's theorem: if \(\sum a_n\) converges, then \(\sum a_n x^n\to\sum a_n\) as \(x\to1^-\).
Let \(S=\sum_{n=0}^\infty a_n\) and \(f(x)=\sum_{n=0}^\infty a_n x^n\) for \(|x|<1\). Let \(R_n=\sum_{k=n}^\infty a_k\). Then \(f(x)=(1-x)\sum_{n=0}^\infty R_n x^n\) (Abel summation). Given \(\epsilon>0\): \(\exists N\) s.t. \(|R_n|<\epsilon\) for \(n\ge N\). For \(x\in(0,1)\): \(|f(x)-S|=|(1-x)\sum R_n x^n|\le(1-x)\sum_{n
Define uniform convergence for a series \(\sum f_n\). State the Cauchy criterion for uniform convergence.
Definition: \(\sum_{n=1}^\infty f_n(x)\) converges uniformly on \(E\) if the partial sums \(S_N(x)=\sum_{n=1}^N f_n(x)\) converge uniformly to \(S(x)\), i.e., \(\sup_{x\in E}|S(x)-S_N(x)|\to0\) as \(N\to\infty\). Cauchy criterion: \(\sum f_n\) converges uniformly on \(E\) iff for every \(\epsilon>0\), \(\exists N\) s.t. for all \(m>n>N\) and all \(x\in E\): \(\left|\sum_{k=n+1}^m f_k(x)\right|<\epsilon\). \(\blacksquare\)
State and prove the Weierstrass M-test for uniform convergence of series.
Statement: If \(|f_n(x)|\le M_n\) for all \(x\in E\) and \(\sum M_n<\infty\), then \(\sum f_n\) converges absolutely and uniformly on \(E\). Proof: For \(m>n\): \(\left|\sum_{k=n+1}^m f_k(x)\right|\le\sum_{k=n+1}^m M_k\). Since \(\sum M_k<\infty\): \(\sum_{k=n+1}^m M_k\to0\) uniformly in \(x\). By the Cauchy criterion, \(\sum f_n\) converges uniformly on \(E\). Absolute convergence: \(\sum|f_n(x)|\le\sum M_n<\infty\). \(\blacksquare\)
Prove: if \(f_n\to f\) uniformly on \([a,b]\), then \(\displaystyle\lim_{n\to\infty}\int_a^b f_n=\int_a^b f\).
Given \(\epsilon>0\): \(\exists N\) s.t. \(|f_n(x)-f(x)|<\dfrac{\epsilon}{b-a}\) for all \(x\in[a,b]\), \(n>N\). Then: \(\left|\int_a^b f_n-\int_a^b f\right|=\left|\int_a^b(f_n-f)\right|\le\int_a^b|f_n-f|\,dx<\dfrac{\epsilon}{b-a}(b-a)=\epsilon\). \(\blacksquare\)
Prove: if each \(f_n\) is continuous on \(E\) and \(f_n\to f\) uniformly, then \(f\) is continuous on \(E\).
Fix \(x_0\in E\), \(\epsilon>0\). By uniform convergence: \(\exists N\) s.t. \(|f_N(x)-f(x)|<\epsilon/3\) for all \(x\). By continuity of \(f_N\): \(\exists\delta>0\) s.t. \(|x-x_0|<\delta\Rightarrow|f_N(x)-f_N(x_0)|<\epsilon/3\). Then: \(|f(x)-f(x_0)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon\). \(\blacksquare\)
Let \(f_n(x)=x^n\) on \([0,1]\). Find the pointwise limit. Is the convergence uniform?
Pointwise: For \(x\in[0,1)\): \(x^n\to0\). For \(x=1\): \(1^n=1\to1\). Limit \(f(x)=\begin{cases}0&x\in[0,1)\\1&x=1\end{cases}\). Not uniform: Each \(f_n\) is continuous but \(f\) is discontinuous at \(x=1\). By the uniform limit theorem, convergence cannot be uniform on \([0,1]\). Alternatively: \(\sup_{x\in[0,1]}|x^n-f(x)|\ge\sup_{x\in[0,1)}x^n=1\not\to0\). \(\blacksquare\)
State and prove Dini's theorem: if \(f_n\to f\) pointwise on compact \(K\), each \(f_n\) continuous, \(f\) continuous, and the convergence is monotone, then it is uniform.
WLOG \(f_n\searrow f\) (decreasing). Let \(g_n=f_n-f\ge0\), \(g_n\searrow0\), each \(g_n\) continuous. Fix \(\epsilon>0\). Let \(U_n=\{x\in K:g_n(x)<\epsilon\}\) — open by continuity. Since \(g_n\to0\): every \(x\in K\) belongs to some \(U_n\). The \(U_n\) are increasing (\(g_n\searrow\Rightarrow U_n\subset U_{n+1}\)) and cover \(K\). By compactness of \(K\): \(K\subset U_{N}\) for some \(N\). So \(g_n(x)<\epsilon\) for all \(x\in K\) and \(n\ge N\). Uniformly. \(\blacksquare\)
If \(f_n\to f\) pointwise and \(f_n'\to g\) uniformly on \([a,b]\), prove \(f'=g\).
Fix \(x_0\in[a,b]\). Write \(f_n(x)=f_n(x_0)+\int_{x_0}^x f_n'(t)\,dt\) (FTC). Taking \(n\to\infty\): LHS \(\to f(x)\); RHS: \(f_n(x_0)\to f(x_0)\), and by uniform convergence \(\int_{x_0}^x f_n'(t)\,dt\to\int_{x_0}^x g(t)\,dt\). So \(f(x)=f(x_0)+\int_{x_0}^x g(t)\,dt\). By FTC: \(f'(x)=g(x)\). \(\blacksquare\)
Prove: a bounded function \(f\) on \([a,b]\) is Riemann integrable iff its set of discontinuities has measure zero (Lebesgue's criterion).
Key idea: The oscillation of \(f\) at \(x\) is \(\omega_f(x)=\lim_{\delta\to0}\sup_{|y-x|,|z-x|<\delta}|f(y)-f(z)|\). The set of discontinuities \(D=\{x:\omega_f(x)>0\}=\bigcup_{n=1}^\infty D_{1/n}\) where \(D_{1/n}=\{x:\omega_f(x)\ge1/n\}\). (\(\Rightarrow\)): If \(f\) is integrable, then for each \(\epsilon,\delta>0\), the partition intervals covering \(D_\epsilon\) have total length \(<\delta\), so \(D_\epsilon\) is measurable of measure zero; hence \(D\) has measure zero. (\(\Leftarrow\)): If \(\mu(D)=0\): given \(\epsilon>0\), cover \(D_{1/n}\) (closed, hence compact) with open intervals of total length \(<\epsilon/(2M)\). The complementary compact set is covered by finitely many intervals where \(\omega_f<1/n\). Construct a partition with \(U-L<\epsilon\). \(\blacksquare\)
Prove that \(f(x)=\begin{cases}1&x\in\mathbb{Q}\\0&x\notin\mathbb{Q}\end{cases}\) is not Riemann integrable on \([0,1]\).
For any partition \(P\) of \([0,1]\): every subinterval \([x_{i-1},x_i]\) contains both rationals and irrationals (by density). So \(M_i=\sup f=1\) and \(m_i=\inf f=0\) on every subinterval. Hence \(U(f,P)=\sum M_i\Delta x_i=1\) and \(L(f,P)=\sum m_i\Delta x_i=0\) for every partition. Since \(U(f,P)-L(f,P)=1\not\to0\), \(f\) fails the Darboux criterion and is not Riemann integrable. \(\blacksquare\)
Define upper and lower Darboux sums. State the Darboux condition for Riemann integrability.
For \(f:[a,b]\to\mathbb{R}\) bounded and partition \(P=\{a=x_0
State and prove both parts of the Fundamental Theorem of Calculus.
FTC I: If \(f\) is Riemann integrable on \([a,b]\) and \(F(x)=\int_a^x f(t)\,dt\), then \(F\) is continuous on \([a,b]\). If \(f\) is continuous at \(x_0\), then \(F'(x_0)=f(x_0)\). Proof: \(|F(x)-F(x_0)|=\left|\int_{x_0}^x f\right|\le M|x-x_0|\to0\). For differentiability at \(x_0\): \(\dfrac{F(x)-F(x_0)}{x-x_0}=\dfrac{1}{x-x_0}\int_{x_0}^x f(t)\,dt\to f(x_0)\) by continuity. FTC II: If \(F'=f\) on \([a,b]\) and \(f\) is Riemann integrable, then \(\int_a^b f=F(b)-F(a)\). Proof: By MVT on each subinterval of a partition: \(\sum f(\xi_i)\Delta x_i=\sum(F(x_i)-F(x_{i-1}))=F(b)-F(a)\). Taking refinements: \(\int_a^b f=F(b)-F(a)\). \(\blacksquare\)
Prove the integration by parts formula: \(\int_a^b u\,dv=uv\big|_a^b-\int_a^b v\,du\), assuming \(u,v\) have continuous derivatives.
By the product rule: \((uv)'=u'v+uv'\). Since \(u,v\in C^1[a,b]\), \((uv)'\) is continuous and integrable. By FTC II: \(\int_a^b(uv)'\,dx=u(b)v(b)-u(a)v(a)=uv\big|_a^b\). So \(\int_a^b u'v\,dx+\int_a^b uv'\,dx=uv\big|_a^b\), giving \(\int_a^b uv'\,dx=uv\big|_a^b-\int_a^b u'v\,dx\). In differential notation with \(dv=v'dx\), \(du=u'dx\): \(\int_a^b u\,dv=uv\big|_a^b-\int_a^b v\,du\). \(\blacksquare\)
State and prove the change of variables formula for Riemann integrals: \(\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)}f(t)\,dt\).
Assume \(g\in C^1[a,b]\), \(f\) continuous. Let \(F(t)=\int_{g(a)}^t f(s)\,ds\), so \(F'=f\) (FTC I). Define \(H(x)=F(g(x))\). By chain rule: \(H'(x)=F'(g(x))\cdot g'(x)=f(g(x))g'(x)\). By FTC II: \(\int_a^b f(g(x))g'(x)\,dx=H(b)-H(a)=F(g(b))-F(g(a))=\int_{g(a)}^{g(b)}f(t)\,dt\). \(\blacksquare\)
Define \(B(p,q)=\int_0^1 x^{p-1}(1-x)^{q-1}\,dx\) for \(p,q>0\). Show it converges and prove \(B(p,q)=B(q,p)\).
Convergence: Singularities at \(x=0\) and \(x=1\). Near \(x=0\): integrand \(\sim x^{p-1}\), integrable since \(p>0\) (\(\int_0^{1/2}x^{p-1}\,dx=\frac{(1/2)^p}{p}<\infty\)). Near \(x=1\): integrand \(\sim(1-x)^{q-1}\), integrable since \(q>0\). Symmetry: Substitute \(t=1-x\): \(B(p,q)=\int_0^1(1-t)^{p-1}t^{q-1}\,dt=B(q,p)\). \(\blacksquare\)
Define \(\Gamma(n)=\int_0^\infty x^{n-1}e^{-x}\,dx\) for \(n>0\). Show \(\Gamma(n+1)=n\Gamma(n)\) and \(\Gamma(1)=1\).
\(\Gamma(1)=1\): \(\int_0^\infty e^{-x}\,dx=[-e^{-x}]_0^\infty=1\). Recurrence: Integrate by parts with \(u=x^n\), \(dv=e^{-x}dx\): \(\Gamma(n+1)=\int_0^\infty x^n e^{-x}\,dx=[-x^n e^{-x}]_0^\infty+n\int_0^\infty x^{n-1}e^{-x}\,dx=0+n\Gamma(n)\). Consequence: For \(n\in\mathbb{N}\): \(\Gamma(n+1)=n!\). The Gamma function extends the factorial to non-integer values. \(\blacksquare\)
Define Lebesgue outer measure on \(\mathbb{R}\). State the properties of measurable sets.
Outer measure: \(\mu^*(E)=\inf\left\{\sum_{k=1}^\infty|I_k|: E\subset\bigcup I_k, I_k\text{ open intervals}\right\}\). Properties of \(\mu^*\): (i) \(\mu^*(\emptyset)=0\); (ii) Monotone: \(E\subset F\Rightarrow\mu^*(E)\le\mu^*(F)\); (iii) Countably subadditive: \(\mu^*(\bigcup E_k)\le\sum\mu^*(E_k)\). Measurability (Carathéodory): \(E\) is (Lebesgue) measurable if for every \(A\subset\mathbb{R}\): \(\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)\). The measurable sets form a \(\sigma\)-algebra, and \(\mu^*\) restricted to them is countably additive. \(\blacksquare\)
Compare the Lebesgue and Riemann integrals. Give a function that is Lebesgue integrable but not Riemann integrable.
Key differences: The Riemann integral is defined for bounded functions on bounded intervals via Darboux sums; requires the set of discontinuities to have measure zero. The Lebesgue integral is defined for a larger class (measurable functions); handles unbounded functions and sets more naturally; dominated convergence and monotone convergence theorems apply. Example: The Dirichlet function \(f=\mathbf{1}_\mathbb{Q}\) on \([0,1]\): not Riemann integrable (shown above), but Lebesgue integrable since \(\mu(\mathbb{Q}\cap[0,1])=0\), so \(\int_{[0,1]}f\,d\mu=1\cdot0=0\). \(\blacksquare\)
State and prove the Monotone Convergence Theorem (MCT) for Lebesgue integrals.
Statement: If \(0\le f_1\le f_2\le\cdots\) are measurable and \(f_n\to f\) pointwise a.e., then \(\int f_n\to\int f\). Proof sketch: Since \(f_n\le f\): \(\int f_n\le\int f\), so \(\limsup\int f_n\le\int f\). For the other inequality: fix \(\alpha\in(0,1)\) and a simple function \(\phi\le f\). Let \(E_n=\{x:f_n(x)\ge\alpha\phi(x)\}\). Then \(E_n\nearrow\) and \(\bigcup E_n\) is (a.e.) the whole space. By continuity of measure: \(\int_{E_n}\phi\to\int\phi\). So \(\int f_n\ge\alpha\int_{E_n}\phi\to\alpha\int\phi\). Letting \(\alpha\to1\) and taking sup over \(\phi\): \(\liminf\int f_n\ge\int f\). \(\blacksquare\)
State and prove Fatou's Lemma: if \(f_n\ge0\) measurable, then \(\int\liminf f_n\le\liminf\int f_n\).
Let \(g_n=\inf_{k\ge n}f_k\). Then \(g_n\le f_n\) so \(\int g_n\le\int f_n\). Also \(g_n\nearrow\liminf f_n\) (pointwise). By MCT: \(\int g_n\to\int\liminf f_n\). Therefore: \(\int\liminf f_n=\lim\int g_n=\liminf\int g_n\le\liminf\int f_n\). \(\blacksquare\)
State and prove the Dominated Convergence Theorem (DCT).
Statement: If \(f_n\to f\) a.e., \(|f_n|\le g\) a.e. with \(g\in L^1\), then \(\int f_n\to\int f\). Proof: Apply Fatou's lemma to \(g+f_n\ge0\) and \(g-f_n\ge0\). For \(g+f_n\): \(\int(g+f)=\int\liminf(g+f_n)\le\liminf\int(g+f_n)=\int g+\liminf\int f_n\). So \(\liminf\int f_n\ge\int f\). For \(g-f_n\): similarly \(\limsup\int f_n\le\int f\). Hence \(\int f_n\to\int f\). \(\blacksquare\)
Prove Hölder's inequality: for \(p,q>1\) with \(1/p+1/q=1\), \(\int|fg|\le\|f\|_p\|g\|_q\).
If \(\|f\|_p=0\) or \(\|g\|_q=0\), both sides are 0. Assume both positive. By Young's inequality: \(ab\le\dfrac{a^p}{p}+\dfrac{b^q}{q}\) for \(a,b\ge0\). Apply with \(a=\dfrac{|f(x)|}{\|f\|_p}\), \(b=\dfrac{|g(x)|}{\|g\|_q}\): \(\dfrac{|f(x)g(x)|}{\|f\|_p\|g\|_q}\le\dfrac{|f(x)|^p}{p\|f\|_p^p}+\dfrac{|g(x)|^q}{q\|g\|_q^q}\). Integrating: \(\dfrac{\int|fg|}{\|f\|_p\|g\|_q}\le\dfrac{1}{p}+\dfrac{1}{q}=1\). \(\blacksquare\)
Prove Minkowski's inequality: for \(p\ge1\), \(\|f+g\|_p\le\|f\|_p+\|g\|_p\).
For \(p=1\): \(\int|f+g|\le\int|f|+\int|g|\). For \(p>1\): \(\int|f+g|^p=\int|f+g|^{p-1}|f+g|\le\int|f+g|^{p-1}|f|+\int|f+g|^{p-1}|g|\). Apply Hölder with exponents \(p\) and \(q=p/(p-1)\): each term \(\le\left(\int|f+g|^p\right)^{1/q}\|f\|_p\) and similarly for \(|g|\). So \(\|f+g\|_p^p\le\|f+g\|_p^{p/q}(\|f\|_p+\|g\|_p)\). Dividing by \(\|f+g\|_p^{p/q}=\|f+g\|_p^{p-1}\) gives \(\|f+g\|_p\le\|f\|_p+\|g\|_p\). \(\blacksquare\)
Show that \(L^2(\mu)\) with inner product \(\langle f,g\rangle=\int fg\,d\mu\) is a Hilbert space (complete inner product space).
Inner product axioms: Linearity in first slot, conjugate symmetry, positivity — all follow from integral properties. \(\langle f,f\rangle=0\Rightarrow\int|f|^2=0\Rightarrow f=0\) a.e. Norm: \(\|f\|=\langle f,f\rangle^{1/2}=\|f\|_2\). Triangle inequality is Minkowski. Completeness (Riesz-Fischer): Let \((f_n)\) be Cauchy in \(L^2\). Extract a subsequence \(\|f_{n_{k+1}}-f_{n_k}\|_2<2^{-k}\). Let \(g=\sum|f_{n_{k+1}}-f_{n_k}|\); by Minkowski \(\|g\|_2\le\sum 2^{-k}=1\). So \(g\in L^2\), finite a.e., and \(f_{n_k}\to f\) a.e. By DCT (or MCT), \(f\in L^2\) and \(f_{n_k}\to f\) in \(L^2\). Since the original is Cauchy, the whole sequence converges to \(f\). \(\blacksquare\)
State Parseval's identity for Fourier series in \(L^2([-\pi,\pi])\): \(\|f\|^2=\sum_{n=-\infty}^\infty|\hat{f}(n)|^2\).
The trigonometric system \(\{e_n(x)=e^{inx}/\sqrt{2\pi}\}_{n\in\mathbb{Z}}\) is an orthonormal basis for \(L^2([-\pi,\pi])\). The Fourier coefficients are \(\hat{f}(n)=\langle f,e_n\rangle=\dfrac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi f(x)e^{-inx}\,dx\). By the general Hilbert space identity (Parseval/Plancherel): if \(\{e_n\}\) is an ONB, then \(\|f\|^2=\sum_n|\langle f,e_n\rangle|^2=\sum_n|\hat{f}(n)|^2\). Equivalently: \(\dfrac{1}{2\pi}\int_{-\pi}^\pi|f|^2\,dx=\sum_{n=-\infty}^\infty|c_n|^2\) where \(c_n=\dfrac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,dx\). \(\blacksquare\)
State the Riesz-Fischer theorem: \(L^p(\mu)\) is complete for \(1\le p\le\infty\).
Statement: For \(1\le p<\infty\): if \((f_n)\) is Cauchy in \(L^p(\mu)\), then there exists \(f\in L^p(\mu)\) with \(\|f_n-f\|_p\to0\). Proof idea: Extract a subsequence with \(\|f_{n_{k+1}}-f_{n_k}\|_p<2^{-k}\). Let \(g_N=\sum_{k=1}^N|f_{n_{k+1}}-f_{n_k}|\). By Minkowski: \(\|g_N\|_p\le1\). By MCT: \(g_N\to g\in L^p\). So \(f_{n_k}=f_{n_1}+\sum_{k=1}^\infty(f_{n_{k+1}}-f_{n_k})\) converges a.e. absolutely to some \(f\). Dominated by \(|f_{n_1}|+g\in L^p\), so by DCT, \(f\in L^p\) and \(f_{n_k}\to f\) in \(L^p\). The Cauchy property gives the full sequence converges to \(f\). \(\blacksquare\)
Define absolute continuity. State: \(F(x)=\int_a^x f\,d\mu\) is absolutely continuous and \(F'=f\) a.e.
Absolute continuity: \(F:[a,b]\to\mathbb{R}\) is absolutely continuous (AC) if for every \(\epsilon>0\), \(\exists\delta>0\) such that for any finite collection of disjoint intervals \((a_k,b_k)\) with \(\sum(b_k-a_k)<\delta\): \(\sum|F(b_k)-F(a_k)|<\epsilon\). Theorem (Lebesgue-Vitali): If \(f\in L^1([a,b])\) then \(F(x)=\int_a^x f\,dt\) is AC. Conversely, if \(F\) is AC then \(F'\) exists a.e., \(F'\in L^1\), and \(F(x)-F(a)=\int_a^x F'\,dt\). This is the Lebesgue version of FTC. AC functions are precisely the antiderivatives of \(L^1\) functions. \(\blacksquare\)
Define total variation. Prove: \(f\in BV[a,b]\) iff \(f=f_1-f_2\) with \(f_1,f_2\) non-decreasing (Jordan decomposition).
Total variation: \(V_a^b(f)=\sup_P\sum|f(x_i)-f(x_{i-1})|\). (\(\Leftarrow\)): Non-decreasing \(\Rightarrow\) BV (shown earlier); differences of BV are BV. (\(\Rightarrow\)): Let \(f_1(x)=V_a^x(f)\) (total variation from \(a\) to \(x\)). Then \(f_1\) is non-decreasing. Set \(f_2=f_1-f\). Claim \(f_2\) is non-decreasing: for \(x
State Helly's selection theorem for sequences of bounded variation functions.
Statement: Let \((f_n)\) be a sequence of functions on \([a,b]\) satisfying: (i) \(|f_n(x)|\le M\) for all \(n,x\) (uniformly bounded); (ii) \(V_a^b(f_n)\le C\) for all \(n\) (uniformly bounded variation). Then there exists a subsequence \((f_{n_k})\) and a function \(f\in BV[a,b]\) such that \(f_{n_k}(x)\to f(x)\) for every \(x\in[a,b]\). Proof sketch: By Jordan decomposition, write each \(f_n=f_n^+-f_n^-\) with \(f_n^{\pm}\) increasing and uniformly bounded. Apply Helly's theorem for monotone functions (diagonal extraction using countable dense sets + monotone limits). \(\blacksquare\)
Prove that the series \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\cdots\) is conditionally convergent but not absolutely convergent.
Not absolutely convergent: \(\sum_{n=1}^\infty\left|\frac{(-1)^{n+1}}{n}\right|=\sum_{n=1}^\infty\frac{1}{n}\) is the harmonic series, which diverges. Conditionally convergent: By the Leibniz alternating series test, let \(b_n=\frac{1}{n}\). Then (i) \(b_n>0\), (ii) \(b_{n+1}
Define a Cauchy sequence in \(\mathbb{R}\). Prove that every convergent sequence is Cauchy, and every Cauchy sequence in \(\mathbb{R}\) is convergent (completeness of \(\mathbb{R}\)).
Cauchy sequence: \((a_n)\) is Cauchy if \(\forall\epsilon>0,\,\exists N\in\mathbb{N}\) such that \(m,n>N\Rightarrow|a_m-a_n|<\epsilon\). Convergent \(\Rightarrow\) Cauchy: Suppose \(a_n\to L\). Given \(\epsilon>0\), choose \(N\) so that \(n>N\Rightarrow|a_n-L|<\epsilon/2\). Then for \(m,n>N\): \[|a_m-a_n|\le|a_m-L|+|L-a_n|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\] Cauchy \(\Rightarrow\) convergent in \(\mathbb{R}\): Every Cauchy sequence is bounded (choose \(N\) for \(\epsilon=1\); then \(|a_n|\le\max(|a_1|,\ldots,|a_N|,|a_{N+1}|+1)\)). By Bolzano–Weierstrass, it has a convergent subsequence \(a_{n_k}\to L\). Using the Cauchy condition: given \(\epsilon>0\), choose \(N\) so \(m,n>N\Rightarrow|a_m-a_n|<\epsilon/2\), and choose \(k\) so \(n_k>N\) and \(|a_{n_k}-L|<\epsilon/2\). Then \(|a_n-L|\le|a_n-a_{n_k}|+|a_{n_k}-L|<\epsilon\). Hence \(a_n\to L\). \(\blacksquare\)
Define the Riemann integral of a bounded function on \([a,b]\). Prove that every continuous function on \([a,b]\) is Riemann integrable.
Riemann integral: For a bounded \(f:[a,b]\to\mathbb{R}\) and partition \(P=\{a=x_0
Use Lagrange's method of multipliers to find the extreme values of \(f(x,y,z)=xyz\) subject to the constraint \(x+y+z=a\) (\(a>0\)). Deduce the AM-GM inequality for three positive numbers.
Lagrange conditions: \(\nabla f=\lambda\nabla g\) where \(g(x,y,z)=x+y+z-a\). So: \((yz,\,xz,\,xy)=\lambda(1,1,1)\). This gives \(yz=xz=xy\). Assuming \(x,y,z>0\): from \(yz=xz\) we get \(y=x\); from \(xz=xy\) we get \(z=y\). So \(x=y=z\). Substituting into \(x+y+z=a\): \(x=y=z=\frac{a}{3}\), giving \(f=\frac{a^3}{27}\). On the boundary (some variable \(\to 0\)), \(f\to 0\), so this critical point is a maximum. AM-GM deduction: Let \(x+y+z=a\) with \(x,y,z>0\). Then: \[xyz\le\left(\frac{a}{3}\right)^3=\left(\frac{x+y+z}{3}\right)^3.\] Taking cube roots: \[(xyz)^{1/3}\le\frac{x+y+z}{3},\] which is the AM-GM inequality for three positive numbers. Equality holds iff \(x=y=z\). \(\blacksquare\)
Using the \(\epsilon\)-\(\delta\) definition, prove that \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x)=x^2\) is continuous at every point. Is it uniformly continuous on \(\mathbb{R}\)?
Continuity at \(x=a\): \(|f(x)-f(a)|=|x^2-a^2|=|x+a||x-a|\). For \(|x-a|<1\): \(|x|\le|a|+1\), so \(|x+a|\le|x|+|a|\le 2|a|+1\). Given \(\epsilon>0\), choose \(\delta=\min\!\left(1,\,\dfrac{\epsilon}{2|a|+1}\right)\). Then \(|x-a|<\delta\) implies: \[|f(x)-f(a)|=(2|a|+1)|x-a|<(2|a|+1)\cdot\frac{\epsilon}{2|a|+1}=\epsilon.\] Not uniformly continuous on \(\mathbb{R}\): Suppose it were. For \(\epsilon=1\), there would exist \(\delta>0\) such that \(|x-y|<\delta\Rightarrow|x^2-y^2|<1\) for all \(x,y\in\mathbb{R}\). But take \(x=n+\delta/2\), \(y=n\) for large \(n\): \(|x-y|=\delta/2<\delta\), yet \(|x^2-y^2|=|x+y|\cdot\delta/2\approx 2n\cdot\delta/2=n\delta\to\infty\). Contradiction. Hence \(f\) is not uniformly continuous on \(\mathbb{R}\). \(\blacksquare\)
Prove that if a sequence \(\{a_n\}\) is monotone increasing and bounded above, then it converges. Find \(\lim_{n\to\infty}\!\left(1+\frac{1}{n}\right)^n\).
Monotone convergence theorem: Let \((a_n)\) be increasing and bounded above. By the completeness of \(\mathbb{R}\), the set \(S=\{a_n:n\in\mathbb{N}\}\) has a supremum \(L=\sup S\). Claim \(a_n\to L\). Given \(\epsilon>0\): \(L-\epsilon\) is not an upper bound of \(S\), so \(\exists N\) with \(a_N>L-\epsilon\). For \(n>N\): \(L-\epsilon
State and prove the Fundamental Theorem of Calculus. If \(F(x)=\int_0^x f(t)\,dt\) where \(f\) is continuous on \([a,b]\), show that \(F'(x)=f(x)\).
Theorem (Part 1): If \(f\) is continuous on \([a,b]\) and \(F(x)=\int_a^x f(t)\,dt\), then \(F\) is differentiable on \((a,b)\) and \(F'(x)=f(x)\). Proof: \[F(x+h)-F(x)=\int_x^{x+h}f(t)\,dt.\] By the Mean Value Theorem for integrals: for each \(h\neq 0\), \(\exists c_h\) between \(x\) and \(x+h\) such that \(\int_x^{x+h}f(t)\,dt=f(c_h)\cdot h\). So \(\frac{F(x+h)-F(x)}{h}=f(c_h)\). As \(h\to 0\), \(c_h\to x\), and by continuity of \(f\): \(f(c_h)\to f(x)\). Hence \(F'(x)=f(x)\). Theorem (Part 2): If \(G'=f\) on \([a,b]\), then \(\int_a^b f(x)\,dx=G(b)-G(a)\). Proof: Let \(H=F-G\) where \(F\) is as above. Then \(H'(x)=f(x)-f(x)=0\) on \((a,b)\). By the MVT, \(H\) is constant, so \(F(x)=G(x)+C\). Evaluating at \(x=a\): \(F(a)=0=G(a)+C\Rightarrow C=-G(a)\). Thus \(\int_a^b f=F(b)=G(b)-G(a)\). \(\blacksquare\)
Test for convergence of the improper integral \(\int_0^\infty\frac{\sin x}{x}\,dx\). Is it absolutely convergent?
Convergence at 0: \(\lim_{x\to 0}\frac{\sin x}{x}=1\), so the integrand extends continuously to \([0,\infty)\); no issue at 0. Convergence at \(\infty\) (Dirichlet's test): Let \(g(x)=\frac{1}{x}\) and \(h(x)=\sin x\). Then \(g\) is positive, monotone decreasing to 0, and \(\left|\int_0^X\sin x\,dx\right|=|{-\cos X+1}|\le 2\) is bounded. By Dirichlet's test, \(\int_0^\infty\frac{\sin x}{x}\,dx\) converges. (Its value is \(\frac{\pi}{2}\).) Not absolutely convergent: \[\int_0^\infty\frac{|\sin x|}{x}\,dx\ge\sum_{n=1}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{(n+1)\pi}\,dx=\sum_{n=1}^\infty\frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin x|\,dx=\frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{n+1}=\infty.\] Hence the integral converges conditionally but not absolutely. \(\blacksquare\)