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 $c-\epsilon <\int _{-m}^{m}f'<c+\epsilon$. By the fundamental theorem of calculus, this gives

$f(-m)+c-\epsilon <f(m)<f(-m)+c+\epsilon$ 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 $\|f_{n}\|_{p}\leq M\!\,$,

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

By Fatou's Lemma,

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

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

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