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.

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