# UMD Analysis Qualifying Exam/Aug07 Real

## Problem 1

 Suppose that $f\!\,$ is a continuous real-valued function with domain $(-\infty ,\infty )\!\,$ and that $f\!\,$ is absolutely continuous on every finite interval $[a,b]\!\,$ . Prove: If $f\!\,$ and $f^{\prime }\!\,$ are both integrable on $(-\infty ,\infty )\!\,$ , then $\int _{-\infty }^{\infty }f^{\prime }=0\!\,$ ### Solution 1

Since $f\!\,$  is absolutely continuous for all $[a,b]\subset R^{1}\!\,$ ,

$\int _{a}^{b}f^{\prime }(x)dx=f(b)-f(a)\!\,$

Hence

$\int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{a,b\rightarrow \infty }\int _{a}^{b}f^{\prime }(x)=\lim _{a,b\rightarrow \infty }[f(b)-f(a)]\!\,$

Since $f^{\prime }\!\,$  is integrable i.e. $f^{\prime }\in L^{1}(-\infty ,\infty )\!\,$ , $\lim _{b\rightarrow \infty }f(b)\!\,$  and $\lim _{a\rightarrow \infty }f(a)\!\,$  exist.

Assume for the sake of contradiction that

$\lim _{b\rightarrow \infty }|f(b)|=\delta >0\!\,$

Then there exists $M\!\,$  such that for all $x>M\!\,$

$|f(x)|>{\frac {\delta }{2}}\!\,$

since $f\!\,$  is continuous. (At some point, $|f|\!\,$  will either monotonically increase or decrease to $\delta \!\,$ .) This implies

{\begin{aligned}\int _{-\infty }^{\infty }|f(x)|dx&\geq \int _{M}^{\infty }|f(x)|dx\\&\geq \int _{M}^{\infty }{\frac {\delta }{2}}dx\\&=\infty \end{aligned}}

which contradicts the hypothesis that $f\!\,$  is integrable i.e. $f\in L^{1}(-\infty ,\infty )\!\,$ . Hence,

$\lim _{b\rightarrow \infty }f(b)=0\!\,$

Using the same reasoning as above,

$\lim _{a\rightarrow -\infty }f(a)=0\!\,$

Hence,

$\int _{-\infty }^{\infty }f^{\prime }(x)dx=\lim _{b\rightarrow \infty }f(b)-\lim _{a\rightarrow -\infty }f(a)=0\!\,$

### Alternate Solution

Suppose $\int _{-\infty }^{\infty }f'=c\neq 0$  (without loss of generality, $c>0$ ). Then for small positive $\epsilon$ , there exists some real $M$  such that for all $m>M$  we have $c-\epsilon <\int _{-m}^{m}f' . By the fundamental theorem of calculus, this gives

$f(-m)+c-\epsilon   for all $m>M$ .

Since $f$  is integrable, this means that for any small positive $\delta$ , there exists an $N$  such that for all $n>N$ , we have $\int _{-\infty }^{-n}+\int _{n}^{\infty }f<\delta$ . But by the above estimate,

$\int _{-\infty }^{-n}f+\int _{n}^{\infty }f>\int _{-\infty }^{-n}f+\int _{-\infty }^{-n}f+(c-\epsilon )=\int _{-\infty }^{-n}2f+\int _{-\infty }^{-n}(c-\epsilon )=\infty$

This contradicts the integrability of $f$ . Therefore, we must have $c=0$ .

## Problem 3

 Suppose that $\{f_{n}\}\!\,$ is a sequence of real valued measurable functions defined on the interval $[0,1]\!\,$ and suppose that $f_{n}(x)\rightarrow f(x)\!\,$ for almost every $x\in [0,1]\!\,$ . Let $p>1\!\,$ and $M>0\!\,$ and suppose that $\|f_{n}\|_{p}\leq M\!\,$ for all $n\!\,$ (a) Prove that $\|f\|_{p}\leq M\!\,$ . (b)Prove that $\|f-f_{n}\|_{1}\rightarrow 0\!\,$ as $n\rightarrow \infty \!\,$ ### Solution 3a

By definition of norm,

$\|f\|_{p}=\left(\int |f(x)|^{p}\,\mathrm {d} x\right)^{\frac {1}{p}}\!\,$

Since $\|f_{n}\|_{p}\leq M\!\,$ ,

$\|f_{n}\|_{p}^{p}\leq M^{p}\!\,$

{\begin{aligned}\|f\|_{p}^{p}&=\int _{0}^{1}|f(x)|^{p}dx\\&=\int _{0}^{1}{\underset {n}{\lim \inf }}|f_{n}(x)|^{p}dx\\&\leq {\underset {n}{\lim \inf }}\int _{0}^{1}|f_{n}(x)|^{p}dx\\&\leq M^{p}\end{aligned}}

which implies, by taking the $p\!\,$ th root,

$\|f\|_{p}\leq M\!\,$

### Solution 3b

By Holder's Inequality, for all $A\subset [0,1]\!\,$  that are measurable,

{\begin{aligned}\int _{A}|(f(x)-f_{n}(x))\cdot 1|dx&\leq \left(\int _{A}|f(x)-f_{n}(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq \left(\int _{A}|2f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2\left(\int _{A}|f(x)|^{p}dx\right)^{\frac {1}{p}}\cdot \left(\int _{A}1^{q}dx\right)^{\frac {1}{q}}\\&\leq 2M\cdot (m(A))^{\frac {1}{q}}\end{aligned}}

where ${\frac {1}{p}}+{\frac {1}{q}}=1\!\,$

Hence, $|f(x)-f_{n}(x)|\leq 2M(m(A))^{\frac {1}{q}}\!\,$

The Vitali Convergence Theorem then implies

$\lim _{n\rightarrow \infty }\int _{0}^{1}|f(x)-f_{n}(x)|dx=\int _{0}^{1}\lim _{n\rightarrow \infty }|f(x)-f_{n}(x)|dx=0\!\,$

## Problem 5

 Suppose $f(x),xf(x)\in L^{2}(R)\!\,$ . Prove that $f(x)\in L^{1}(R)\!\,$ and that $\|f\|_{1}\leq {\sqrt {2}}(\|f\|_{2}+\|xf\|_{2})\!\,$ ### Solution 5

{\begin{aligned}\int _{R}|f(x)|dx&=\int _{|x|>1}|xf(x|\cdot {\frac {1}{|x|}}dx+\int _{|x|<1}|f(x)|\cdot 1dx\\&\leq \left(\int _{|x|>1}|xf(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|>1}{\frac {1}{|x|^{2}}}dx\right)^{\frac {1}{2}}+\left(\int _{|x|<1}|f(x)|^{2}dx\right)^{\frac {1}{2}}\cdot \left(\int _{|x|<1}1^{2}dx\right)^{\frac {1}{2}}\end{aligned}}