Problem 1 (problem1). Let $E$ be a normed linear space. Show that $E$ is complete if and only if, whenever $\sum_{1}^{\infty}\left\|x_{n}\right\|<\infty,$ then $\sum_{1}^{\infty} x_{n}$ converges to an $s \in E .$
Problem 2 (problem2). Let f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}\) be analytic and one-to-one on $|z|<1 .$ Suppose that $|f(z)|<1$ for all $|z|<1$ (a) Prove that $$\sum_{n=1}^{\infty} n\left|a_{n}\right|^{2} \leq 1$$ (b) Is the constant 1 the best possible?
Problem 3 (problem3). Let $f$ be integrable on the real line with respect to Lebesgue measure. Evaluate $\lim _{n \rightarrow \infty} \int_{-\infty}^{\infty} f(x-n)\left(\frac{x}{1+|x|}\right) d x$ Justify all steps.
Problem 4 (problem4). Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral $$\int_{0}^{1} \int_{y}^{1} x^{-3 / 2} \cos \left(\frac{\pi y}{2 x}\right) d x d y$$
Problem 5 (problem5). Let $m$ be Lebesgue measure on the real line $\mathbb{R},$ and for each Lebesgue measurable subset $E$ of $\mathbb{R}$ define $$\mu(E)=\int_{E} \frac{1}{1+x^{2}} d m(x)$$ Show that $m$ is absolutely continuous with respect to $\mu,$ and compute the Radon-Nikodym derivative $d m / d \mu$.
Proof (problem1). Solution: Suppose $E$ is complete. Let $\left\{x_{n}\right\} \subset E$ be absolutely convergent; i.e., $\sum\left\|x_{n}\right\|<\infty$. We must $$\sum_{n=1}^{\infty} x_{n}:=\lim _{N \rightarrow \infty} \sum_{n=1}^{N} x_{n}=s \in E$$ Let $S_{N}=\sum_{n=1}^{N} x_{n} .$ Then, for any $j \in \mathbb{N}$ $$\left\|S_{N+j}-S_{N}\right\|=\left\|\sum_{n=N+1}^{N+j} x_{n}\right\| \leq \sum_{n=N+1}^{N+j}\left\|x_{n}\right\| \rightarrow 0$$ as $N \rightarrow \infty,$ since $\sum\left\|x_{n}\right\|<\infty .$ Therefore, $\left\{S_{N}\right\}$ is a Cauchy sequence. Since $E$ is complete, there is an $s \in E$ such that $\sum_{n=1}^{\infty} x_{n}=\lim _{N \rightarrow \infty} S_{N}=s$ Conversely, suppose whenever $\sum_{1}^{\infty}\left\|x_{n}\right\|<\infty,$ then $\sum_{1}^{\infty} x_{n}$ converges to an $s \in E .$ Let $\left\{y_{n}\right\} \subset E$ be a Cauchy sequence. That is, $\left\|y_{n}-y_{m}\right\| \rightarrow 0$ as $n, m \rightarrow \infty$. Let $n_{1}<n_{2}<\cdots$ be=”” a=”” subsequence=”” such=”” that=”” $$=”” n,=”” m=”” \geq=”” n_{j}=”” \rightarrow\left\|y_{n}-y_{m}\right\|<2^{-j}=”” next=”” observe,=”” for=”” $k=””>1$$$ y_{n_{k}}=y_{n_{1}}+\left(y_{n_{2}}-y_{n_{1}}\right)+\left(y_{n_{3}}-y_{n_{2}}\right)+\cdots+\left(y_{n_{k}}-y_{n_{k-1}}\right)=y_{n_{1}}+\sum_{j=1}^{k-1}\left(y_{n_{j+1}}-y_{n_{j}}\right) $$and$$ \sum_{j=1}^{\infty}\left\|y_{n_{j+1}}-y_{n_{j}}\right\|<\sum_{j=1}^{\infty} 2^{-j}=1 $$By hypothesis, this implies that$$ y_{n_{k}}-y_{n_{1}}=\sum_{j=1}^{k-1}\left(y_{n_{j+1}}-y_{n_{j}}\right) \rightarrow s \in E $$as $k \rightarrow \infty$. We have thus found a subsequence $\left\{y_{n_{k}}\right\} \subseteq\left\{y_{n}\right\}$ having a limit in $E$. Finally, since $\left\{y_{n}\right\}$ is Cauchy, it is quite easy to verify that $\left\{y_{n}\right\}$ must converge to the same limit. This proves that every Cauchy sequence in $E$ converges to a point in $E$. </n_{2}<\cdots\)> Proof (problem2). (a) This is a special case of the following area theorem: Suppose $f(z)=\sum_{n=0}^{\infty} a_{n} z^{n}$ is a holomorphic function which maps the unit disk $D=\{|z|<1\}$ bijectively onto a domain $f(D)=G$ having area $A$. Then$$ A=\pi \sum_{n=1}^{\infty} n\left|a_{n}\right|^{2} $$Proof: The area of the image of $D$ under $f$ is the integral over $D$ of the Jacobian of $f$. That is,$$ A=\iint_{D}\left|f^{\prime}(z)\right|^{2} d x d y $$Compute $\left|f^{\prime}(z)\right|$ by differentiating the power series of $f(z)$ term by term.$$ f^{\prime}(z)=\sum_{n=1}^{\infty} n a_{n} z^{n-1} $$Next, take the squared modulus,$$ \left|f^{\prime}(z)\right|^{2}=\sum_{m, n=1}^{\infty} m n a_{m} \bar{a}_{n} z^{m-1} \bar{z}^{n-1} $$This gives,$$ A=\iint_{D} \sum_{m, n=1}^{\infty} m n a_{m} \bar{a}_{n} z^{m-1} \bar{z}^{n-1} d x d y $$Letting $z=r e^{i \theta}$,$$ A=\sum_{m, n=1}^{\infty} m n a_{m} \bar{a}_{n} \int_{0}^{1} \int_{0}^{2 \pi} r^{m+n-1} e^{i(m-n) \theta} d \theta d r $$Now, for all $k \neq 0$, the integral of $e^{i k \theta}$ over $0 \leq \theta<2 \pi$ vanishes, so the only non-vanishing terms of the series are those for which $m=n$. That is.$$ A=2 \pi \sum_{n=1}^{\infty} n^{2}\left|a_{n}\right|^{2} \int_{0}^{1} r^{2 n-1} d r=\pi \sum_{n=1}^{\infty} n^{2}\left|a_{n}\right|^{2} $$To apply this theorem to the problem at hand, note that the hypotheses of the problem imply that $f$ maps the unit disk bijectively onto its range $f(D),$ which is contained inside $D$ and, therefore, has area less or equal to $\pi .$ This and (38) together imply$$ \pi \geq \pi \sum_{n=1}^{\infty} n^{2}\left|a_{n}\right|^{2} $$which gives the desired inequality. (b) The identity function $f(z)=z$ satisfies the given hypotheses and its power series expansion has coefficients $a_{1}=1$ and $0=a_{0}=a_{2}=a_{3}=\cdots .$ This shows that the upper bound of 1 is obtained and is therefore the best possible. Proof (problem3). Fix $n>0$. Consider the change of variables, $y=x-n$. Then $d y=d x$ and $x=y+n$, so$$ \begin{aligned} \int_{-\infty}^{\infty} f(x-n) \frac{x}{1+|x|} d x &=\int_{-\infty}^{\infty} f(y) \frac{y+n}{1+|y+n|} d y \\ &=\int_{-n}^{\infty} f(y) \frac{y+n}{1+y+n} d y+\int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)} d y \end{aligned} $$Note that, when $y \geq-n, \frac{y+n}{1+y+n} \in[0,1),$ and increases to 1 as $n$ tends to infinity. Thus,$$ 0 \leq|f(y)| \frac{y+n}{1+y+n} \leq|f(y)| $$for all $y>-n .$ Define the function $^{14}$$$ g_{n}(y)=f(y) \frac{y+n}{1+y+n} \mathbf{1}_{[-n, \infty)}(y) $$Then $\left|g_{n}\right| \leq|f|$ and $\lim _{n \rightarrow \infty} g_{n}=f .$ Therefore, by the dominated convergence theorem,$$ \lim _{n \rightarrow \infty} \int_{-n}^{\infty} f(y) \frac{y+n}{1+y+n} d y=\lim _{n \rightarrow \infty} \int_{-\infty}^{\infty} g_{n}(y) d y=\int_{-\infty}^{\infty} f(y) d y $$Next, consider the second term in (7). Define the function$$ h_{n}(y)=f(y) \frac{y+n}{1-(y+n)} \mathbf{1}_{(-\infty,-n]}(y) $$It is not hard to check that$$ \frac{|y+n|}{|1-(y+n)|} \mathbf{1}_{(-\infty,-n]}(y) \in[0,1) $$from which it follows that $\left|h_{n}\right| \leq|f| .$ Also, it is clear that, for all $y$,$$ \lim _{n \rightarrow \infty} h_{n}(y)=f(y) \lim _{n \rightarrow \infty} \frac{y+n}{1-(y+n)} \mathbf{1}_{(-\infty,-n]}(y)=0 $$Therefore, the dominated convergence theorem implies that$$ \lim _{n \rightarrow \infty} \int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)} d y=0 $$Combining the two results above, we see that $\lim _{n \rightarrow \infty} \int_{-\infty}^{\infty} f(x-n)\left(\frac{x}{1+|x|}\right) d x=\int_{-\infty}^{\infty} f(x) d x$ Proof (problem4). By Tonelli’s theorem, if $f(x, y) \geq 0$ is measurable and one of the iterated integrals $\iint f(x, y) d x d y$ or $\iint f(x, y) d y d x$ exists, then they both exist and are equal. Moreover, if one of the iterated integrals is finite, then $f(x, y) \in L^{1}(d x, d y) .$ Fubini’s theorem states: if $f(x, y) \in L^{1}(d x, d y),$ then the iterated integrals exist and are equal. Now let $g(x, y)=x^{-3 / 2} \cos (\pi y / 2 x),$ and apply the Tonelli theorem to the non-negative measurable function $|g(x, y)|$ as follows:$$ \int_{0}^{1} \int_{0}^{x}|g(x, y)| d y d x=\int_{0}^{1} \int_{0}^{x}|x|^{-3 / 2}\left|\cos \left(\frac{\pi y}{2 x}\right)\right| d y d x \leq \int_{0}^{1} \int_{0}^{x} x^{-3 / 2} \cdot 1 d y d x=\int_{0}^{1} x^{-1 / 2} d x=2 $$Thus one of the iterated integrals of $|g(x, y)|$ is finite which, by the Tonelli theorem, implies $g(x, y) \in L^{1}(d x, d y)$. Therefore, the Fubini theorem applies to $g(x, y),$ and gives the first of the following equalities:$$ \begin{aligned} \int_{0}^{1} \int_{y}^{1} x^{-3 / 2} \cos \left(\frac{\pi y}{2 x}\right) d x d y &=\int_{0}^{1} \int_{0}^{x} x^{-3 / 2} \cos \left(\frac{\pi y}{2 x}\right) d y d x \\ &=\int_{0}^{1} x^{-3 / 2} \cdot \frac{2 x}{\pi}\left[\sin \left(\frac{\pi y}{2 x}\right)\right]_{y=0}^{y=x} d x \\ &=\int_{0}^{1} \frac{2}{\pi} x^{-1 / 2} d x \\ &=\frac{2}{\pi}\left[2 x^{1 / 2}\right]_{x=0}^{x=1} \\ &=\frac{4}{\pi} \end{aligned} $$Proof (problem5). Obviously both measures are non-negative. We must first prove $m \ll \mu .$ To this end, suppose $m(E)>$ $0,$ where $E \in \mathfrak{M},$ the $\sigma$ -algebra of Lebesgue measurable sets. Then, if we can show $\mu(E)>0,$ this will establish that the implication $\mu(E)=0 \Rightarrow m(E)=0$ holds for all $E \in \mathfrak{M} ;$ i.e., $m \ll \mu$. For $n=1,2 \ldots .$ define$$ A_{n}=\left\{x \in \mathbb{R}: \frac{1}{n+1}<\frac{1}{1+x^{2}} \leq \frac{1}{n}\right\} $$Then $A_{i} \cap A_{j}=\emptyset$ for all $i \neq j$ in $\mathbb{N},$ and, for all $n=1,2, \ldots$$$ \mu\left(A_{n}\right) \geq \frac{1}{n+1} m\left(A_{n}\right) $$Also, $\mathbb{R}=\cup A_{n},$ since $0<\frac{1}{1+x^{2}} \leq 1$ holds for all $x \in \mathbb{R}$. Therefore,$$ \mu(E)=\mu\left(E \cap\left(\cup_{n} A_{n}\right)\right)=\mu\left(\cup_{n}\left(A_{n} \cap E\right)\right)=\sum_{n} \mu\left(A_{n} \cap E\right) $$The last equality might need a bit of justification: Since $f(x)=\frac{1}{1+x^{2}}$ is continuous, hence measurable, the sets $\left\{A_{n}\right\}$ are measurable. Therefore, the last equality holds by countable additivity of disjoint measurable sets. Now note that $m(E)=\sum m\left(A_{n} \cap E\right)>0$ implies the existence of an $n \in \mathbb{N}$ such that $m\left(A_{n} \cap E\right)>0 .$ Therefore,$$ \mu(E) \geq \mu\left(A_{n} \cap E\right) \geq \frac{1}{n+1} m\left(A_{n} \cap E\right)>0 $$which proves that $m \ll \mu .$ By the Radon-Nikodym theorem (A.1), there is a unique $h \in L^{1}(\mu)$ such that$$ m(E)=\int h d \mu, \quad \text { and } \quad \int f d m=\int f h d \mu \quad \forall f \in L^{1}(m) $$In particular, if $E \in \mathfrak{M}$ and $f(x)=\frac{1}{1+x^{2}} \chi_{E},$ then$$ \mu(E)=\int_{E} \frac{1}{1+x^{2}} d m(x)=\int_{E} \frac{h(x)}{1+x^{2}} d \mu(x) $$That is, $\int_{E} d \mu=\int_{E} \frac{h(x)}{1+x^{2}} d \mu(x)$ holds for all measurable sets $E,$ which implies $^{16}$ that, $\frac{h(x)}{1+x^{2}}=1$ holds for $\mu$ -almost every $x \in \mathbb{R} .$ Therefore,$$ \frac{d m}{d \mu}(x)=h(x)=1+x^{2}  One final note: $h$ is uniquely defined only up to an equivalence class of functions that are equal to $1+x^{2}, \mu$ -a.e.