## Fall 2018.1 #real_analysis/qual/completed

Let $$f(x) = \frac 1 x$$. Show that $$f$$ is uniformly continuous on $$(1, \infty)$$ but not on $$(0,\infty)$$.

• Uniform continuity: \begin{align*} \forall \varepsilon>0, \exists \delta({\varepsilon})>0 \quad\text{such that}\quad {\left\lvert {x-y} \right\rvert}<\delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon} .\end{align*}

• Negating uniform continuity: $$\exists {\varepsilon}> 0$$ such that $$\forall \delta({\varepsilon})$$ there exist $$x, y$$ such that $${\left\lvert {x-y} \right\rvert} < \delta$$ and $${\left\lvert {f(x) - f(y)} \right\rvert} > {\varepsilon}$$.

• Archimedean property: for all $$x,y\in {\mathbf{R}}$$ there exists an $$n \in {\mathbb{N}}$$ such that $$nx>y$$. Take $$x={\varepsilon}, y=1$$, so $$n{\varepsilon}> 1$$ and $${1\over n} < {\varepsilon}$$.

1 is the only constant around, so try to use it for uniform continuity. To negate, find a bad $$x$$: since $$1/x$$ blows up near zero, go hunting for small $$x$$s!

• Claim: $$f(x) = \frac 1 x$$ is uniformly continuous on $$(c, \infty)$$ for any $$c > 0$$.
• Note that \begin{align*} {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert} > c > 0 \implies {\left\lvert {xy} \right\rvert} = {\left\lvert {x} \right\rvert}{\left\lvert {y} \right\rvert} > c^2 \implies \frac{1}{{\left\lvert {xy} \right\rvert}} < \frac 1 {c^{2}} .\end{align*}

• Letting $$\varepsilon$$ be arbitrary, choose $$\delta < \varepsilon c^2$$.

• Note that $$\delta$$ does not depend on $$x, y$$.
• Then \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} &= {\left\lvert {\frac 1 x - \frac 1 y} \right\rvert} \\ &= \frac{{\left\lvert {x-y} \right\rvert}}{xy} \\ &\leq \frac{\delta}{xy} \\ &< \frac{\delta}{c^2} \\ &< \varepsilon .\end{align*}

• Claim: $$f$$ is not uniformly continuous when $$c=0$$.
• Take $$\varepsilon < 1$$, and let $$\delta = \delta({\varepsilon})$$ be arbitrary.
• Let $$x_n = \frac 1 n$$ for $$n\geq 1$$.
• Choose $$n$$ large enough such that $${1\over n} < \delta$$
• Then a computation: \begin{align*} {\left\lvert {x_n - x_{n+1}} \right\rvert} &= \frac 1 n - \frac 1 {n+1} \\ &= {1\over n(n+1) } \\ &< {1\over n} \\ &< \delta ,\end{align*}
• Why this can be done: by the Archimedean property of $${\mathbf{R}}$$, for any $$\delta\in {\mathbf{R}}$$, one can choose choose $$n$$ such that $$n\delta > 1$$. We’ve also used that $$n+1 > 1$$ so $${1\over n+1}< 1$$
• Note that $$f(x_n) = n$$, so \begin{align*} {\left\lvert {f(x_{n+1}) - f(x_{n})} \right\rvert} = (n+1) - n = 1 > \varepsilon .\end{align*}

## Fall 2017.1 #real_analysis/qual/completed

Let \begin{align*} f(x) = \sum _{n=0}^{\infty} \frac{x^{n}}{n !}. \end{align*} Describe the intervals on which $$f$$ does and does not converge uniformly.

• $$f_N\to f$$ uniformly $$\iff$$ $${\left\lVert {f_N - f} \right\rVert}_\infty \to 0$$.
• Applied to sums: \begin{align*} \sum_{0 \leq k\leq N} f_n \overset{u}\to \sum_{k\geq 0} f_n \iff {\left\lVert {\sum_{k\geq N+1} f_n } \right\rVert}_{\infty} \to 0 .\end{align*}
• An infinite sum is defined as the pointwise limit of its partial sums: \begin{align*} \sum_{n=0}^\infty c_n x^n \coloneqq\lim_{N\to \infty} \sum_{n=0}^N c_n x^n .\end{align*}
• Uniformly decaying terms for uniformly convergent series: if $$\sum_{n=0}^\infty f_n(x)$$ converges uniformly on a set $$A$$, then \begin{align*} {\left\lVert {f_n} \right\rVert}_{\infty, A} \coloneqq\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}
• $$M{\hbox{-}}$$test: if $$f_n:A \to{\mathbf{C}}$$ with $${\left\lVert {f_n} \right\rVert}_\infty < M_n$$ and $$\sum M_n < \infty$$, then $$\sum f_n$$ converges uniformly and absolutely.
• If the $$f_n$$ are continuous, the uniform limit theorem implies $$\sum f_n$$ is also continuous.

No real place to start, so pick the nicest place: compact intervals. Then bounded intervals, then unbounded sets.

• Set $$f_N(x) = \sum_{n=1}^N {x^n \over n!}$$.

• Then by definition, $$f_N(x) \to f(x)$$ pointwise on $${\mathbf{R}}$$.
• Claim: $$f_N$$ converges on compact intervals

• For any compact interval $$[-M, M]$$, we have \begin{align*} {\left\lVert {f_N(x) - f(x)} \right\rVert}_\infty &= \sup_{x\in [-M, M] } ~{\left\lvert {\sum_{n=N+1}^\infty {x^n \over {n!}} } \right\rvert} \\ &\leq \sup_{x\in [-M, M] } ~\sum_{n=N+1}^\infty {\left\lvert { {x^n \over {n!}} } \right\rvert} \\ &\leq \sum_{n=N+1}^\infty {M^n \over n!} \\ &\leq \sum_{n=0}^\infty {M^n \over {n!} } \quad\text{since all additional terms are positive} \\ &= e^M \\ &<\infty ,\end{align*} so $$f_N \to f$$ uniformly on $$[-M, M]$$ by the M-test.
• Note: we’ve used that this power series converges to $$e^x$$ pointwise everywhere.
• This argument shows that $$f$$ converges on any bounded set.

• Claim: $$f_N$$ does not converge uniformly on all of $${\mathbf{R}}$$.

• Uniformly convergent sums have uniformly decaying terms: \begin{align*} \sum_{n\leq N} g_n \overset{N\to\infty}\longrightarrow\sum g_n \text{ uniformly on } A \implies {\left\lVert {g_n} \right\rVert}_{\infty, A} \coloneqq\sup_{x\in A} {\left\lvert {g_n(x)} \right\rvert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}

• Take $$B_N$$ a ball of radius $$N$$ about 0, then for $$N>1$$, note that $$x=N$$ on the boundary and so \begin{align*} {\left\lVert {x^k \over k!} \right\rVert}_{\infty, B_N} = {N^k \over k!} \overset{N\to\infty}\longrightarrow\infty .\end{align*}

• Conclusion: $$f_N$$ converges on any bounded $$A\subseteq {\mathbf{R}}$$ but not on all of $${\mathbf{R}}$$.

## Spring 2017.4 #real_analysis/qual/completed

Let $$f(x, y)$$ on $$[-1, 1]^2$$ be defined by \begin{align*} f(x, y) = \begin{cases} \frac{x y}{\left(x^{2}+y^{2}\right)^{2}} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases} \end{align*} Determine if $$f$$ is integrable.

• Just Calculus.
• $$1/r$$ is not integrable on $$(0, 1)$$.

Switching to polar coordinates and integrating over the quarter of the unit disc $$D \cap Q_1 \subseteq I^2$$ in quadrant 1, we have \begin{align*} \int_{I^2} f \, dA &\geq \int_D f \, dA \\ &= \int_0^{\pi/2} \int_0^1 \frac{r^2 \cos(\theta)\sin(\theta)}{r^4} ~r~\,dr\,d\theta\\ &= \int_0^{\pi/2} \int_0^1 \frac{\cos(\theta)\sin(\theta)}{r} \,dr\,d\theta\\ &= \qty{ \int_0^1 {1\over r } \,dr} \qty{ \int_0^{\pi/2} \cos(\theta)\sin(\theta) \,d\theta} \\ &= \qty{ \int_0^1 {1\over r } \,dr} \qty{ \int_0^{1} u \,du} && u=\sin(\theta)\\ &= {1\over 2}\qty{ \int_0^1 {1\over r } \,dr} \\ &\longrightarrow\infty .\end{align*}

## Fall 2014.1 #real_analysis/qual/completed

Let $$\left\{{f_n}\right\}$$ be a sequence of continuous functions such that $$\sum f_n$$ converges uniformly.

Prove that $$\sum f_n$$ is also continuous.

• The uniform limit theorem.
• $${\varepsilon}/3$$ trick.

If $$F_N\to F$$ uniformly with each $$F_N$$ continuous, then $$F$$ is continuous.

• Follows from an $$\varepsilon/3$$ argument: \begin{align*} {\left\lvert {F(x) - F(y} \right\rvert} \leq {\left\lvert {F(x) - F_N(x)} \right\rvert} + {\left\lvert {F_N(x) - F_N(y)} \right\rvert} + {\left\lvert {F_N(y) - F(y)} \right\rvert} \leq {\varepsilon}\to 0 .\end{align*}

• The first and last $${\varepsilon}/3$$ come from uniform convergence of $$F_N\to F$$.
• The middle $${\varepsilon}/3$$ comes from continuity of each $$F_N$$.
• Now setting $$F_N\coloneqq\sum_{n=1}^N f_n$$ yields a finite sum of continuous functions, which is continuous.
• Each $$F_N$$ is continuous and $$F_N\to F$$ uniformly, so $$F$$ is continuous.

## Spring 2015.1 #real_analysis/qual/completed

Let $$(X, d)$$ and $$(Y, \rho)$$ be metric spaces, $$f: X\to Y$$, and $$x_0 \in X$$.

Prove that the following statements are equivalent:

• For every $$\varepsilon > 0 \quad \exists \delta > 0$$ such that $$\rho( f(x), f(x_0) ) < \varepsilon$$ whenever $$d(x, x_0) < \delta$$.
• The sequence $$\left\{{f(x_n)}\right\}_{n=1}^\infty \to f(x_0)$$ for every sequence $$\left\{{x_n}\right\} \to x_0$$ in $$X$$.

• What it means for a sequence to converge.
• Trading $$N$$s for $$\delta$$s.

• Let $$\left\{{x_n}\right\} \overset{n\to\infty}\to x_0$$ be arbitrary; we want to show $$\left\{{f(x_n)}\right\}\overset{n\to\infty}\to f(x_0)$$.
• We thus want to show that for every $${\varepsilon}>0$$, there exists an $$N({\varepsilon})$$ such that \begin{align*}n\geq N({\varepsilon}) \implies \rho(f(x_n), f(x_0)) < {\varepsilon}.\end{align*}
• Let $${\varepsilon}>0$$ be arbitrary, then by (1) choose $$\delta$$ such that $$\rho(f(x), f(x_0)) < {\varepsilon}$$ when $$d(x, x_0) < \delta$$.
• Since $$x_n\to x$$, there is some $$N$$ such that $$n\geq N \implies d(x_n, x_0) < \delta$$
• Then for $$n\geq N$$, $$d(x_n, x_0) < \delta$$ and thus $$\rho(f(x_n), f(x_0)) < {\varepsilon}$$, so $$f(x_n)\to f(x_0)$$ by definition.

The direct implication is not a good idea here, since you need a handle on all $$x$$ in a neighborhood of $$x_0$$, not just a specific sequence.

• By contrapositive, show that $$\not 1\implies \not 2$$.
• Need to show: if $$f$$ is not $${\varepsilon}{\hbox{-}}\delta$$ continuous at $$x_0$$, then there exists a sequence $$x_n\to x_0$$ where $$f(x_n)\not\to f(x_0)$$.
• Negating $$1$$, we have that there exists an $${\varepsilon}>0$$ such that for all $$\delta$$, there exists an $$x$$ with $$d(x, x_0) < \delta$$ but $$\rho(f(x), f(x_0))>{\varepsilon}$$
• So take a sequence of deltas $$\delta_n = {1\over n}$$, apply this to produce a sequence $$x_n$$ with $$d(x_n, x_0) < \delta_n \coloneqq{1\over n} \longrightarrow 0$$ and $$\rho(f(x_n), f(x_0)) > {\varepsilon}$$ for all $$n$$.
• This yields a sequence $$x_n \to x_0$$ where $$f(x_n) \not\to f(x_0)$$.

## Fall 2014.2 #real_analysis/qual/completed

Let $$I$$ be an index set and $$\alpha: I \to (0, \infty)$$.

• Show that \begin{align*} \sum_{i \in I} a(i):=\sup _{\substack{ J \subset I \\ J \text { finite }}} \sum_{i \in J} a(i)<\infty \implies I \text{ is countable.} \end{align*}

• Suppose $$I = {\mathbf{Q}}$$ and $$\sum_{q \in \mathbb{Q}} a(q)<\infty$$. Define \begin{align*} f(x):=\sum_{\substack{q \in \mathbb{Q}\\ q \leq x}} a(q). \end{align*} Show that $$f$$ is continuous at $$x \iff x\not\in {\mathbf{Q}}$$.

• Can always filter sets $$X$$ with a function $$X\to {\mathbf{R}}$$.
• Countable union of countable sets is still countable.
• Continuity: $$\lim_{y\to x} f(y) = f(x)$$ from either side.
• Trick: pick enumerations of countable sets and reindex sums

• Set $$S \coloneqq\sum_{i\in I} \alpha(i)$$, we will show that $$S<\infty \implies I$$ is countable.
• Write \begin{align*} I = \bigcup_{n\geq 0} S_n, && S_n \coloneqq\left\{{i\in I {~\mathrel{\Big\vert}~}\alpha(i) \geq {1\over n}}\right\} .\end{align*}
• Note that $$S_n \subseteq S$$ for all $$n$$, so $$\sum_{i\in I}\alpha(i) \geq \sum_{i\in S_n} \alpha(i)$$ for all $$n$$.
• It suffices to show that $$S_n$$ is countable, since $$I$$ is a countable union of $$S_n$$.
• There is an inequality \begin{align*} \infty &> S \coloneqq\sum_{i\in I} \alpha(i) \\ &\geq \sum_{i\in S_n} \alpha(i) \\ &\geq \sum_{i\in S_n} {1\over n} \\ &= {1\over n} \sum_{i\in S_n} 1 \\ &= \qty{1\over n} # S_n \\ \\ \implies \infty &> n S \geq # S_n .\end{align*}

• We’ll prove something more general: let $$Q = \left\{{q_k}\right\}$$ be countable and $$\left\{{\alpha_k \coloneqq\alpha(q_k)}\right\}$$ be summable, and define \begin{align*} f(x) \coloneqq\sum_{q_k\leq x} \alpha_k .\end{align*}

• $$f$$ is always discontinuous precisely on the countable set $$Q$$ and continuous on $${\mathbf{R}}\setminus Q$$.

• $$f$$ is always left-continuous, is right-continuous at $$x\in{\mathbf{R}}\setminus Q$$, and not right-continuous at $$x\in Q$$

• $$f$$ has jump discontinuities at every $$q_m$$, where the jump is precisely $$\alpha_m$$.

• This follows from computing the left and right limits: \begin{align*} f(x^+) &= \lim_{h\to 0} \sum_{q_k \leq x+h} \alpha_k = \sum_{q_k\leq x} \alpha_k = \sum_{q_k < x} \alpha_k + \sum_{q_k = x} \alpha_k \\ f(x^-) &= \lim_{h\to 0} \sum_{q_k \leq x-h} \alpha_k = \sum_{q_k < x} \alpha_k ,\end{align*} where we’ve used that $$\left\{{q_k \leq x}\right\} = \left\{{q_k < x}\right\} {\textstyle\coprod}\left\{{x}\right\}$$ in the first equality.

• Then if $$x=q_m$$ for some $$m$$, \begin{align*} f(x^+) &= f(q_m^+) = \sum_{q_k < q_m} \alpha_k + \alpha_m \\ f(x^-) &= f(a_m^-) = \sum_{q_k< q_m} \alpha_k ,\end{align*} which clearly differ if $$\alpha_m \neq 0$$.

• Taking $$x\not\in Q$$, we have $$\left\{{q_k \leq x}\right\} = \left\{{q_k < x}\right\}$$, since $$\left\{{q_k=x}\right\} = \emptyset$$, so \begin{align*} f(x^+) &= \sum_{q_k\leq x} \alpha_k = \sum_{q_k < x} \alpha_k \\ f(x^-) &= \sum_{q_k< x} \alpha_k ,\end{align*} so the limits agree.

• To recover the result in the problem, let $${\mathbf{Q}}= \left\{{q_k}\right\}$$ be any enumeration of the rationals.

# General Analysis

## Fall 2021.1 #real_analysis/qual/completed

Let $$\left\{x_{n}\right\}_{n-1}^{\infty}$$ be a sequence of real numbers such that $$x_{1}>0$$ and \begin{align*} x_{n+1}=1-\left(2+x_{n}\right)^{-1}=\frac{1+x_{n}}{2+x_{n}} \text {. } \end{align*} Prove that the sequence $$\left\{x_{n}\right\}$$ converges, and find its limit.

If a limit $$L$$ exists, we have $$x_n\to L$$ for all $$n$$, so \begin{align*} L = {1+L\over 2+L} \implies L^2 + L - 1 = 0 \implies L = -{1\over 2}\qty{-1 \pm \sqrt 5} .\end{align*} Noting that $$\sqrt{5} > 1$$, the condition $$x_1>0$$ and a small induction noting that if $$x_n>0$$ then $${1+x_n \over 2+x_n}>0$$, the only solution can be $$L = -1 + \sqrt 5$$. To see that this does converge, write $$f(z) = 1 - (2+z)^{-1}$$ so that $$x_{n+1} = f(x_n)$$. The claim is that $$f$$ is a contracting map on a metric space, which implies it has a unique fixed point $$z_0$$ by the Banach fixed point theorem, and if $$f(z_0) = z_0$$ then $$z_0 = L$$. This follows from the mean value theorem, since \begin{align*} {\left\lvert {f(z) - f(w)} \right\rvert} = {\left\lvert {f'(\xi)} \right\rvert}{\left\lvert {z-w} \right\rvert} < {\left\lvert {z-w} \right\rvert} && \text{for some } \xi \in (z, w) .\end{align*} Since $$f'(z) = (2+z)^{-2}$$ satisfies $$0 < f'(z) < 1$$ for all $$z$$, we have \begin{align*} {\left\lvert {f(z) - f(w)} \right\rvert} \leq {\left\lvert {z-w} \right\rvert} .\end{align*}

## Fall 2020.1 #real_analysis/qual/completed

Show that if $$x_n$$ is a decreasing sequence of positive real numbers such that $$\sum_{n=1}^\infty x_n$$ converges, then \begin{align*} \lim_{n\to\infty} n x_n = 0. \end{align*}

See this MSE post for many solutions: https://math.stackexchange.com/questions/4603/if-a-n-subset0-infty-is-non-increasing-and-sum-a-n-infty-then-lim Note that the “obvious” thing here is fiddly: there are bounds on the slices \begin{align*} (N-M \pm 1) x_N \leq \sum_{M\leq k \leq N} a_k \leq (N-M\pm 1) x_M ,\end{align*} but arranging it so that the constants match the indices in $$(N-M \pm 1)x_N \approx Nx_N$$ requires something clever.

Fix $${\varepsilon}>0$$, we’ll find $$n\gg 1$$ so that $$nx_n < {\varepsilon}$$. Find $$n, m$$ with $$n>m$$ large enough so that \begin{align*} {\varepsilon}> \sum_{m+1\leq k \leq n} x_k \geq \sum_{m+1\leq k \leq n}x_n = (m-n)x_n .\end{align*} Then rearrange: \begin{align*} {\varepsilon}> (m-n)x_n \implies nx_n < {\varepsilon}+ mx_n .\end{align*} Now choose $$n$$ large enough so that $$x_n < {\varepsilon}$$, which holds since $$\sum x_n < \infty$$, to obtain \begin{align*} nx_n < {\varepsilon}+ m{\varepsilon}= {\varepsilon}(1+m) \to 0 .\end{align*}

## Spring 2020.1 #real_analysis/qual/completed

Prove that if $$f: [0, 1] \to {\mathbf{R}}$$ is continuous then \begin{align*} \lim_{k\to\infty} \int_0^1 kx^{k-1} f(x) \,dx = f(1) .\end{align*}

• DCT
• Weierstrass Approximation Theorem
• If $$f: [a, b] \to {\mathbf{R}}$$ is continuous, then for every $${\varepsilon}>0$$ there exists a polynomial $$p_{\varepsilon}(x)$$ such that $${\left\lVert {f - p_{\varepsilon}} \right\rVert}_\infty < {\varepsilon}$$.

• Suppose $$p$$ is a polynomial, then integrate by parts: \begin{align*} \lim_{k\to\infty} \int_0^1 kx^{k-1} p(x) \, dx &= \lim_{k\to\infty} \int_0^1 \qty{ {\frac{\partial }{\partial x}\,}x^k } p(x) \, dx \\ &= \lim_{k\to\infty} \left[ x^k p(x) \Big|_0^1 - \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx \right] \quad\text{IBP}\\ &= p(1) - \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx ,\end{align*}

• Thus it suffices to show that \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,} (x) } \, dx = 0 .\end{align*}

• Integrating by parts a second time yields \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx &= \lim_{k\to\infty} {x^{k+1} \over k+1} {\frac{\partial p}{\partial x}\,}(x) \Big|_0^1 - \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2 p}{\partial x^2}\,}(x)} \, dx \\ &= \lim_{k\to\infty} {p'(1) \over k+1} - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \int_0^1 \lim_{k\to\infty} {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \quad\text{by DCT} \\ &= - \int_0^1 0 \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= 0 .\end{align*}

• The DCT can be applied here because polynomials are smooth and $$[0, 1]$$ is compact, so $${\frac{\partial ^2 p}{\partial x^2}\,}$$ is bounded on $$[0, 1]$$ by some constant $$M$$ and \begin{align*} \int_0^1 {\left\lvert {x^k {\frac{\partial ^2 p}{\partial x^2}\,} (x)} \right\rvert} \leq \int_0^1 1\cdot M = M < \infty.\end{align*}
• So the result holds when $$f$$ is a polynomial.

• Now use the Weierstrass approximation theorem:

• If $$f: [a, b] \to {\mathbf{R}}$$ is continuous, then for every $${\varepsilon}>0$$ there exists a polynomial $$p_{\varepsilon}(x)$$ such that $${\left\lVert {f - p_{\varepsilon}} \right\rVert}_\infty < {\varepsilon}$$.
• Thus \begin{align*} {\left\lvert { \int_0^1 kx^{k-1} p_{\varepsilon}(x)\,dx - \int_0^1 kx^{k-1}f(x)\,dx } \right\rvert} &= {\left\lvert { \int_0^1 kx^{k-1} \qty{p_{\varepsilon}(x) - f(x)} \,dx } \right\rvert} \\ &\leq {\left\lvert { \int_0^1 kx^{k-1} {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \,dx } \right\rvert} \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \cdot {\left\lvert { \int_0^1 kx^{k-1} \,dx } \right\rvert} \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \cdot x^k \Big|_0^1 \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \\ \\ &\overset{{\varepsilon}\to 0}\to 0 \end{align*}

and the integrals are equal.

• By the first argument, \begin{align*}\int_0^1 kx^{k-1} p_{\varepsilon}(x) \,dx = p_{\varepsilon}(1) \text{ for each } {\varepsilon}\end{align*}

• Since uniform convergence implies pointwise convergence, $$p_{\varepsilon}(1) \overset{{\varepsilon}\to 0}\to f(1)$$.

## Fall 2019.1 #real_analysis/qual/completed

Let $$\{a_n\}_{n=1}^\infty$$ be a sequence of real numbers.

• Prove that if $$\displaystyle\lim_{n\to \infty } a_n = 0$$, then \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{1}+\cdots+a_{n}}{n}=0 \end{align*}

• Prove that if $$\displaystyle\sum_{n=1}^{\infty} \frac{a_{n}}{n}$$ converges, then \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{1}+\cdots+a_{n}}{n}=0 \end{align*}

• Cesaro mean/summation.
• Break series apart into pieces that can be handled separately.
• Idea: once $$N$$ is large enough, $$a_k \approx S$$, and all smaller terms will die off as $$N\to \infty$$.

• Prove a stronger result: \begin{align*} a_k \to S \implies S_N\coloneqq\frac 1 N \sum_{k=1}^N a_k \to S .\end{align*}

• For any $${\varepsilon}> 0$$, use convergence $$a_k \to S$$: choose (and fix) $$M = M({\varepsilon})$$ large enough such that \begin{align*} k\geq M+1 \implies {\left\lvert {a_k - S} \right\rvert} < \varepsilon .\end{align*}

• With $$M$$ fixed, choose $$N = N(M, {\varepsilon})$$ large enough so that $${1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} < {\varepsilon}$$.

• Then \begin{align*} \left|\left(\frac{1}{N} \sum_{k=1}^{N} a_{k}\right)-S\right| &= {1\over N} {\left\lvert { \qty{\sum_{k=1}^N a_k} - NS } \right\rvert} \\ &= {1\over N} {\left\lvert { \qty{\sum_{k=1}^N a_k} - \sum_{k=1}^N S } \right\rvert} \\ &=\frac{1}{N}\left|\sum_{k=1}^{N}\left(a_{k}-S\right)\right| \\ &\leq \frac{1}{N} \sum_{k=1}^{N}\left|a_{k}-S\right| \\ &= {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + \sum_{k=M+1}^N {\left\lvert {a_k - S} \right\rvert} \\ &\leq {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + \sum_{k=M+1}^N {{\varepsilon}} \quad \text{since } a_k \to S\\ &= {1\over N} \sum_{k=1}^{M} {\left\lvert {a_k - S} \right\rvert} + (N - M){{\varepsilon}} \\ &\leq {\varepsilon}+ (N(M, {\varepsilon}) - M({\varepsilon})){\varepsilon} .\end{align*}

\todo[inline]{Revisit, not so clear that the last line can be made smaller than ${\varepsilon}$, since $M, N$ both depend on ${\varepsilon}$...}

• Define \begin{align*} \Gamma_n \coloneqq\sum_{k=n}^\infty \frac{a_k}{k} .\end{align*}

• $$\Gamma_1 = \sum_{k=1}^n \frac{ a_k } k$$ is the original series and each $$\Gamma_n$$ is a tail of $$\Gamma_1$$, so by assumption $$\Gamma_n \overset{n\to\infty}\to 0$$.

• Compute \begin{align*} \frac 1 n \sum_{k=1}^n a_k &= \frac 1 n (\Gamma_1 + \Gamma_2 + \cdots + \Gamma_{n} \mathbf{- \Gamma_{n+1}}) \\ .\end{align*}

• This comes from consider the following summation:

• Use part (a): since $$\Gamma_n \overset{n\to\infty}\to 0$$, we have $${1\over n} \sum_{k=1}^n \Gamma_k \overset{n\to\infty}\to 0$$.

• Also a minor check: $$\Gamma_n \to 0 \implies {1\over n}\Gamma_n \to 0$$.

• Then \begin{align*} \frac 1 n \sum_{k=1}^n a_k &= \frac 1 n (\Gamma_1 + \Gamma_2 + \cdots + \Gamma_{n} \mathbf{- \Gamma_{n+1}}) \\ &= \qty{ {1\over n } \sum_{k=0}^n \Gamma_k } - \qty{{1\over n}\Gamma_{n+1} } \\ &\overset{n\to\infty}\to 0 .\end{align*}

## Fall 2018.4 #real_analysis/qual/completed

Let $$f\in L^1([0, 1])$$. Prove that \begin{align*} \lim_{n \to \infty} \int_{0}^{1} f(x) {\left\lvert {\sin n x} \right\rvert} ~d x= \frac{2}{\pi} \int_{0}^{1} f(x) ~d x \end{align*}

Hint: Begin with the case that $$f$$ is the characteristic function of an interval.

\todo[inline]{Ask someone to check the last approximation part.}

• Converting floor/ceiling functions to inequalities: $$x-1 \leq {\left\lfloor x \right\rfloor} \leq x$$.

Case of a characteristic function of an interval $$[a, b]$$:

• First suppose $$f(x) = \chi_{[a, b]}(x)$$.

• Note that $$\sin(nx)$$ has a period of $$2\pi/n$$, and thus $${\left\lfloor (b-a) \over (2\pi / n) \right\rfloor} = {\left\lfloor n(b-a)\over 2\pi \right\rfloor}$$ full periods in $$[a, b]$$.

• Taking the absolute value yields a new function with half the period

• So $${\left\lvert {\sin(nx)} \right\rvert}$$ has a period of $$\pi/n$$ with $${\left\lfloor n(b-a) \over \pi \right\rfloor}$$ full periods in $$[a, b]$$.
• We can compute the integral over one full period (which is independent of which period is chosen)

• We can use translation invariance of the integral to compute this over the period $$0$$ to $$\pi/n$$.
• Since $$\sin(nx)$$ is positive, it equals $${\left\lvert {\sin(nx)} \right\rvert}$$ on its first period, so we have \begin{align*} \int_{\text{One Period}} {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \int_0^{\pi/n} \sin(nx)\,dx \\ &= {1\over n} \int_0^\pi \sin(u) \,du \quad u = nx \\ &= {1\over n} \qty{-\cos(u)\mathrel{\Big|}_0^\pi} \\ &= {2 \over n} .\end{align*}
• Then break the integral up into integrals over full periods $$P_1, P_2, \cdots, P_N$$ where $$N \coloneqq{\left\lfloor n(b-a)/\pi \right\rfloor}$$

• Noting that each period is of length $$\pi\over n$$, so letting $$L_n$$ be the regions falling outside of a full period, we have

• Thus \begin{align*} \int_a^b {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \qty{ \sum_{j=1}^{N} \int_{P_j} {\left\lvert {\sin(nx)} \right\rvert} \, dx } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \qty{ \sum_{j=1}^{N} {2\over n} } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= N \qty{2\over n} + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &\coloneqq{\left\lfloor (b-a) n \over \pi \right\rfloor} {2\over n} + R_n \\ &\coloneqq(b-a)C_n + R_n \end{align*} where (claim) $$C_n \overset{n\to\infty}\to {2\over \pi}$$ and $$R(n) \overset{n\to\infty}\to 0$$.

• $$C_n \to {2\over \pi}$$: \begin{align*} {n-1 \over n} \qty{2\over \pi} = {n-1 \over \pi} \qty{2\over n} \leq {\left\lfloor n\over \pi \right\rfloor}\qty{2\over n} \leq {n \over \pi}\qty{2\over n} = {2 \over \pi} ,\end{align*} then use the fact that $${n-1 \over n} \to 1$$.

• Then equality follows by the Squeeze theorem.
• $$R_n \to 0$$:

• We use the fact that $$m(L_n) \to 0$$, then $$\int_{L_n} {\left\lvert {\sin(nx)} \right\rvert} \leq \int_{L_n} 1 = m(L_n) \to 0$$.
• This follows from the fact that $$L_n$$ is the complement of $$\cup_j P_j$$, the set of full periods, so \begin{align*} m(L_n) &= m(b-a) - \sum m(P_j) \\ &= \qty{b-a} - {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\overset{n\to \infty}\to (b-a) - (b-a) \\ &= 0 .\end{align*} where we’ve used the fact that \begin{align*} \qty{\pi \over n} \qty{(b-a)n-1 \over \pi} &\leq {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\leq \qty{\pi \over n} \qty{(b-a)n\over \pi} \\ &= (b-a) ,\end{align*} then taking $$n\to \infty$$ sends the LHS to $$b-a$$, forcing the middle term to be $$b-a$$ by the Squeeze theorem.

General case:

• By linearity of the integral, the result holds for simple functions:

• If $$f = \sum c_j \chi_{E_j}$$ where $$E_j = [a_j, b_j]$$, we have \begin{align*} \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx &= \int_0^1 \sum c_j \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j \int_0^1 \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j (b_j - a_j) {2\over \pi} \\ &= {2\over \pi} \sum c_j (b_j - a_j) \\ &= {2\over \pi} \sum c_j m(E_j) \\ &\coloneqq{2\over \pi} \int_0^1 f .\end{align*}
• Since $$f\in L^1$$, where simple functions are dense, choose $$s_n\nearrow f$$ where $${\left\lVert {s_N - f} \right\rVert}_1 < {\varepsilon}$$, then \begin{align*} {\left\lvert { \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert} \,dx - \int_0^1 s_N(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx } \right\rvert} &= {\left\lvert { \int_0^1 \qty{f(x) - s_N(x)} {\left\lvert {\sin(nx)} \right\rvert} \,dx } \right\rvert} \\ &\leq \int_0^1 {\left\lvert { f(x) - s_N(x)} \right\rvert} {\left\lvert {\sin(nx)} \right\rvert} \,dx \\ &= {\left\lVert { \qty{f - s_N} {\left\lvert {\sin(nx)} \right\rvert} } \right\rVert}_1 \\ &\leq {\left\lVert {f-s_N} \right\rVert}_1 \cdot {\left\lVert {{\left\lvert {\sin(nx)} \right\rvert}} \right\rVert}_\infty \quad\text{by Holder}\\ &\leq {\varepsilon}\cdot 1 ,\end{align*}

• So the integrals involving $$s_N$$ converge to the integral involving $$f$$, and \begin{align*} \lim_{n\to\infty} \int f(x){\left\lvert {\sin(nx)} \right\rvert} &= \lim_{n\to\infty} \lim_{N\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \\ &= \lim_{N\to\infty} \lim_{n\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \quad\text{because ?}\\ &= \lim_{N\to \infty} {2\over \pi} \int s_N(x) \\ &= {2\over \pi} \int f ,\end{align*} which is the desired result.

## Fall 2017.4 #real_analysis/qual/completed

Let \begin{align*} f_{n}(x) = n x(1-x)^{n}, \quad n \in {\mathbb{N}}. \end{align*}

• Show that $$f_n \to 0$$ pointwise but not uniformly on $$[0, 1]$$.

• Show that \begin{align*} \lim _{n \to \infty} \int _{0}^{1} n(1-x)^{n} \sin x \, dx = 0 \end{align*}

Hint for (a): Consider the maximum of $$f_n$$.

• $$\sum f_n < \infty \iff \sup f_n \to 0$$.
• Negating uniform convergence: $$f_n\not\to f$$ uniformly iff $$\exists {\varepsilon}$$ such that $$\forall N({\varepsilon})$$ there exists an $$x_N$$ such that $${\left\lvert {f(x_N) - f(x)} \right\rvert} > {\varepsilon}$$.
• Exponential inequality: $$1+y \leq e^y$$ for all $$y\in {\mathbf{R}}$$.

$$f_n\to 0$$ pointwise:

• Finding the maximum: can check that $${\frac{\partial f_n}{\partial x}\,} = x(1-x)^{n-1} \qty{1 + (n^2-1)x}$$
• This has critical points $$x=0, 1, {-1 \over n^2 + 1}$$, and the latter is a global max on $$[0, 1]$$.
• Set $$x_n \coloneqq{-1 \over n^2 + 1}$$
• Compute \begin{align*} \lim f_n(x_n) = \lim_{n\to \infty } {-n \over n^2 + 1} \qty{1 + x_n}^n = 0\cdot 1 = 0 .\end{align*}
• So $$\sup f_n \to 0$$, forcing $$f_n \to 0$$ pointwise.

The convergence is not uniform:

• Let $$x_n = \frac 1 n$$ and $$\varepsilon > e^{-1}$$, then \begin{align*} {\left\lVert {nx(1-x)^n - 0} \right\rVert}_\infty &\geq {\left\lvert {nx_n (1-x_n)^n} \right\rvert} \\ &= {\left\lvert {\left( 1 - \frac 1 n\right)^n} \right\rvert} \\ &> e^{-1} \\ &> \varepsilon .\end{align*}

• Here we’ve used that $$(1 + {x\over n})^n \leq e^x$$ for all $$x\in {\mathbf{R}}$$ and all $$n$$.
• Follows from $$1+y \leq e^y$$ applied to $$y = x/n$$.
• Thus $${\left\lVert {f_n - 0} \right\rVert}_\infty = {\left\lVert {f_n} \right\rVert}_\infty > e^{-1} > 0$$.

• ?
\todo[inline]{Possible to use part a with $\sin(x) \leq x$ on $[0, \pi/2]$?}
• Noting that $$\sin(x) \leq 1$$, we have \begin{align*} {\left\lvert {\int_0^1 n(1-x)^{n} \sin(x)} \right\rvert} &\leq \int_0^1 {\left\lvert {n(1-x)^n \sin(x)} \right\rvert} \\ &\leq \int_0^1 {\left\lvert {n (1-x)^n} \right\rvert} \\ &= n\int_0^1 (1-x)^n \\ &= -\frac{n(1-x)^{n+1}}{n+1} \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

## Spring 2017.3 #real_analysis/qual/work

Let \begin{align*} f_{n}(x) = a e^{-n a x} - b e^{-n b x} \quad \text{ where } 0 < a < b. \end{align*}

Show that

• $$\sum_{n=1}^{\infty} \left|f_{n}\right|$$ is not in $$L^{1}([0, \infty), m)$$

Hint: $$f_n(x)$$ has a root $$x_n$$.

• \begin{align*} \sum_{n=1}^{\infty} f_{n} \text { is in } L^{1}([0, \infty), m) {\quad \operatorname{and} \quad} \int _{0}^{\infty} \sum _{n=1}^{\infty} f_{n}(x) \,dm = \ln \frac{b}{a} \end{align*}
\todo[inline]{Not complete.}
\todo[inline]{Walk through.}
• $$f_n$$ has a root: \begin{align*} ae^{-nax} = be^{-nbx} &\iff {1\over n} = e^{-nbx} e^{nax} = e^{n(b-a)x} \iff x = {\ln\qty{a\over b} \over n(a-b)} \coloneqq x_n .\end{align*}

• Thus $$f_n$$ only changes sign at $$x_n$$, and is strictly positive on one side of $$x_n$$.

• Then \begin{align*} \int_{\mathbf{R}}\sum_n {\left\lvert {f_n(x)} \right\rvert}\,dx &= \sum_n \int_{\mathbf{R}}{\left\lvert {f_n(x)} \right\rvert} \,dx \\ &\geq \sum_n \int_{x_n}^\infty f_n(x) \, dx \\ &= \sum_n {1\over n} \qty{ e^{-bnx} - e^{-anx}\Big|_{x_n}^\infty } \\ &= \sum_n {1\over n} \qty{ e^{-bnx_n} - e^{-anx_n} } .\end{align*}

?

## Fall 2016.1 #real_analysis/qual/completed

Define \begin{align*} f(x) = \sum_{n=1}^{\infty} \frac{1}{n^{x}}. \end{align*} Show that $$f$$ converges to a differentiable function on $$(1, \infty)$$ and that \begin{align*} f'(x) =\sum_{n=1}^{\infty}\left(\frac{1}{n^{x}}\right)^{\prime}. \end{align*}

Hint: \begin{align*} \left(\frac{1}{n^{x}}\right)' = -\frac{1}{n^{x}} \ln n \end{align*}

• ?
• Set $$f_N(x) \coloneqq\sum_{n=1}^N n^{-x}$$, so $$f(x) = \lim_{N\to\infty} f_N(x)$$.

• If an interchange of limits is justified, we have \begin{align*} {\frac{\partial }{\partial x}\,} \lim_{N\to\infty} \sum_{n=1}^N n^{-x} &= \lim_{h\to 0} \lim_{N\to\infty} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &\mathop{\mathrm{=}}_{?} \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &= \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ {\sum_{n=1}^N n^{-x}} - {n^{-(x+h)} }\right] \quad\text{(1)} \\ &= \lim_{N\to\infty} \sum_{n=1}^N \lim_{h\to 0} {1\over h} \left[ n^{-x} - n^{-(x+h)} \right] \quad\text{since this is a finite sum} \\ &\coloneqq\lim_{N\to\infty} \sum_{n=1}^N {\frac{\partial }{\partial x}\,}\qty{1 \over n^x} \\ &= \lim_{N\to\infty} \sum_{n=1}^N -{\ln(n) \over n^x} ,\end{align*} where the combining of sums in (1) is valid because $$\sum n^{-x}$$ is absolutely convergent for $$x>1$$ by the $$p{\hbox{-}}$$test.

• Thus it suffices to justify the interchange of limits and show that the last sum converges on $$(1, \infty)$$.

• Claim: $$\sum n^{-x}\ln(n)$$ converges.

• Use the fact that for any fixed $${\varepsilon}>0$$, \begin{align*} \lim_{n\to\infty} {\ln(n) \over n^{\varepsilon}} \mathop{\mathrm{=}}^{L.H.} \lim_{n\to\infty}{1/n \over {\varepsilon}n^{{\varepsilon}-1}} = \lim_{n\to\infty} {1\over {\varepsilon}n^{\varepsilon}} = 0 ,\end{align*}

• This implies that for a fixed $${\varepsilon}>0$$ and for any constant $$c>0$$ there exists an $$N$$ large enough such that $$n\geq N$$ implies $$\ln(n)/n^{\varepsilon}< c$$, i.e. $$\ln(n) < c n^{{\varepsilon}}$$.

• Taking $$c=1$$, we have $$n\geq N \implies \ln(n) < n^{\varepsilon}$$

• We thus break up the sum: \begin{align*} \sum_{n\in {\mathbb{N}}} {\ln(n) \over n^x} &= \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {\ln(n) \over n^x} \\ &\leq \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {n^{\varepsilon}\over n^x} \\ &\coloneqq C_{\varepsilon}+ \sum_{n=N}^\infty {n^{\varepsilon}\over n^x} \quad \text{with $C_{\varepsilon}<\infty$ a constant}\\ &= C_{\varepsilon}+ \sum_{n=N}^\infty {1 \over n^{x-{\varepsilon}}} ,\end{align*} where the last term converges by the $$p{\hbox{-}}$$test if $$x-{\varepsilon}> 1$$.

• But $${\varepsilon}$$ can depend on $$x$$, and if $$x\in (1, \infty)$$ is fixed we can choose $${\varepsilon}< {\left\lvert {x-1} \right\rvert}$$ to ensure this.

• Claim: the interchange of limits is justified.

\todo[inline]{?}

## Fall 2016.5 #real_analysis/qual/completed

Let $$\phi\in L^\infty({\mathbf{R}})$$. Show that the following limit exists and satisfies the equality \begin{align*} \lim _{n \to \infty} \left(\int _{\mathbb{R}} \frac{|\phi(x)|^{n}}{1+x^{2}} \, dx \right) ^ {\frac{1}{n}} = {\left\lVert {\phi} \right\rVert}_\infty. \end{align*}

• ?

Let $$L$$ be the LHS and $$R$$ be the RHS.

Claim: $$L\leq R$$.

• Since $${\left\lvert {\phi } \right\rvert}\leq {\left\lVert {\phi} \right\rVert}_\infty$$ a.e., we can write 
\begin{align*} L^{1\over n} &\coloneqq\int_{\mathbf{R}}{ {\left\lvert {\phi(x)} \right\rvert}^n \over 1+ x^2} \\ &\leq \int_{\mathbf{R}}{ {\left\lVert {\phi} \right\rVert}_\infty^n \over 1+ x^2} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \int_{\mathbf{R}}{1\over 1 + x^2} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \arctan(x)\Big|_{-\infty}^{\infty} \\ &= {\left\lVert {\phi} \right\rVert}_\infty^n \qty{{\pi \over 2} - {-\pi \over 2} } \\ &= \pi {\left\lVert {\phi} \right\rVert}_\infty^n \\ \\ \implies L^{1\over n} &\leq \sqrt[n]{\pi {\left\lVert {\phi} \right\rVert}_\infty^n} \\ \implies L &\leq \pi^{1\over n} {\left\lVert {\phi} \right\rVert}_\infty \\ &\overset{n\to \infty }\to {\left\lVert {\phi} \right\rVert}_\infty ,\end{align*} {=html} where we've used the fact that $c^{1\over n} \overset{n\to\infty}\to 1$ for any constant $c$. \todo[inline]{Actually true? Need conditions?}{=tex}

Claim: $$R\leq L$$.

• We will show that $$R\leq L + {\varepsilon}$$ for every $${\varepsilon}>0$$.
• Set \begin{align*} S_{\varepsilon}\coloneqq\left\{{x\in {\mathbf{R}}^n{~\mathrel{\Big\vert}~}{\left\lvert {\phi(x)} \right\rvert} \geq {\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}\right\} .\end{align*}
• Then we have \begin{align*} \int_{\mathbf{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx &\geq \int_{S_{\varepsilon}} {{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx \quad S_{\varepsilon}\subset {\mathbf{R}}\\ &\geq \int_{S_{\varepsilon}} { \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n \over 1 +x^2}\,dx \qquad\text{by definition of }S_{\varepsilon}\\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n \int_{S_{\varepsilon}} { 1 \over 1 +x^2}\,dx \\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n C_{\varepsilon}\qquad\text{where $C_{\varepsilon}$ is some constant} \\ \\ \implies \qty{ \int_{\mathbf{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx }^{1\over n} &\geq \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}} C_{\varepsilon}^{1 \over n} \\ &\overset{n\to\infty}\to \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}} \cdot 1 \\ &\overset{{\varepsilon}\to 0}\to {\left\lVert {\phi} \right\rVert}_\infty ,\end{align*} where we’ve again used the fact that $$c^{1\over n} \to 1$$ for any constant.

## Fall 2016.6 #real_analysis/qual/completed

Let $$f, g \in L^2({\mathbf{R}})$$. Show that \begin{align*} \lim _{n \to \infty} \int _{{\mathbf{R}}} f(x) g(x+n) \,dx = 0 \end{align*}

\todo[inline]{Rewrite solution.}

• Cauchy Schwarz: $${\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {g} \right\rVert}_1$$.
• Small tails in $$L^p$$.

• Use the fact that $$L^p$$ has small tails: if $$h\in L^2({\mathbf{R}})$$, then for any $${\varepsilon}> 0$$, \begin{align*} \forall {\varepsilon},\, \exists N\in {\mathbb{N}}{\quad \operatorname{such that} \quad}\int_{{\left\lvert {x} \right\rvert} \geq {N}} {\left\lvert {h(x)} \right\rvert}^2 \,dx < {\varepsilon} .\end{align*}

• So choose $$N$$ large enough so that \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {g(x)} \right\rvert}^2 < {\varepsilon}\\ \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {f(x)} \right\rvert}^2 < {\varepsilon}\\ .\end{align*}

• Then write \begin{align*} \int_{{\mathbf{R}}^d} f(x) g(x+n) \,dx = \int_{{\left\lVert {x} \right\rVert} \leq N} f(x)g(x+n)\,dx + \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx .\end{align*}

• Bounding the second term: apply Cauchy-Schwarz \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx \leq \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {f(x)} \right\rvert}^2}^{1\over 2} \cdot \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \leq {\varepsilon}^{1\over 2} \cdot {\left\lVert {g} \right\rVert}_2 .\end{align*}

• Bounding the first term: also Cauchy-Schwarz, after variable changes \begin{align*} \int_{{\left\lVert {x} \right\rVert} \leq N} f(x) g(x+n)\,dx &= \int_{-N}^N f(x) g(x+n)\,dx \\ &= \int_{-N+n}^{N+n} f(x-n) g(x)\,dx \\ &\leq \int_{-N+n}^{\infty} f(x-n) g(x)\,dx \\ &\leq \qty{\int_{-N+n}^{\infty} {\left\lvert {f(x-n)} \right\rvert}^2}^{1\over 2}\cdot \qty{\int_{-N+n}^{\infty} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \\ &\leq {\left\lVert {f} \right\rVert}_2 \cdot {\varepsilon}^{1\over 2} .\end{align*}

• Then as long as $$n\geq 2N$$, we have \begin{align*} \int {\left\lvert {f(x) g(x+n)} \right\rvert} \leq \qty{{\left\lVert {f} \right\rVert}_2 + {\left\lVert {g} \right\rVert}_2} \cdot {\varepsilon}^{1\over 2} .\end{align*}

## Spring 2016.1 #real_analysis/qual/work

For $$n\in {\mathbb{N}}$$, define \begin{align*} e_{n} = \left (1+ {1\over n} \right)^{n} {\quad \operatorname{and} \quad} E_{n} = \left( 1+ {1\over n} \right)^{n+1} \end{align*}

Show that $$e_n < E_n$$, and prove Bernoulli’s inequality: \begin{align*} (1+x)^n \geq 1+nx && -1 < x < \infty ,\,\, n\in {\mathbb{N}} .\end{align*}

Use this to show the following:

• The sequence $$e_n$$ is increasing.
• The sequence $$E_n$$ is decreasing.
• $$2 < e_n < E_n < 4$$.
• $$\lim _{n \to \infty} e_{n} = \lim _{n \to \infty} E_{n}$$.

## Fall 2015.1 #real_analysis/qual/work

Define \begin{align*} f(x)=c_{0}+c_{1} x^{1}+c_{2} x^{2}+\ldots+c_{n} x^{n} \text { with } n \text { even and } c_{n}>0. \end{align*}

Show that there is a number $$x_m$$ such that $$f(x_m) \leq f(x)$$ for all $$x\in {\mathbf{R}}$$.

## Spring 2014.2 #real_analysis/qual/completed

Let $$\left\{{a_n}\right\}$$ be a sequence of real numbers such that \begin{align*} \left\{{b_n}\right\} \in \ell^2({\mathbb{N}}) \implies \sum a_n b_n < \infty. \end{align*} Show that $$\sum a_n^2 < \infty$$.

Note: Assume $$a_n, b_n$$ are all non-negative.

\todo[inline]{Have someone check!}

• Define a sequence of operators \begin{align*} T_N: \ell^2 &\to \ell^1\\ \left\{{b_n}\right\} &\mapsto \sum_{n=1}^N a_n b_n .\end{align*}

• By assumption, these are well defined: the image is $$\ell^1$$ since $${\left\lvert {T_N(\left\{{b_n}\right\})} \right\rvert} < \infty$$ for all $$N$$ and all $$\left\{{b_n}\right\} \in \ell^2$$.

• So each $$T_N \in \qty{\ell^2} {}^{ \vee }$$ is a linear functional on $$\ell^2$$.

• For each $$x\in \ell^2$$, we have $${\left\lVert {T_N(x)} \right\rVert}_{{\mathbf{R}}} = \sum_{n=1}^N a_n b_n < \infty$$ by assumption, so each $$T_N$$ is pointwise bounded.

• By the Uniform Boundedness Principle, $$\sup_N {\left\lVert {T_N} \right\rVert}_{\text{op}} < \infty$$.

• Define $$T = \lim_{N \to\infty } T_N$$, then $${\left\lVert {T} \right\rVert}_{\text{op}} < \infty$$.

• By the Riesz Representation theorem, \begin{align*} \sqrt{\sum a_n^2} \coloneqq{\left\lVert {\left\{{a_n}\right\}} \right\rVert}_{\ell^2} = {\left\lVert {T} \right\rVert}_{\qty{\ell^2} {}^{ \vee }} = {\left\lVert {T} \right\rVert}_{\text{op}} < \infty .\end{align*}

• So $$\sum a_n^2 < \infty$$.

#real_analysis/qual/completed #todo #real_analysis/qual/work