UMD Analysis Qualifying Exam/Aug07 Real

Problem 1

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

Solution 1

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

$\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)dx=f(b)-f(a)$

Hence

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}^{\prime }(x)dx=\underset{a,b\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)=\underset{a,b\to \mathrm{\infty }}{\lim }[f(b)-f(a)]$

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

Assume for the sake of contradiction that

$\underset{b\to \mathrm{\infty }}{\lim }|f(b)|=\delta >0$

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

$\begin{array}{cc}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|f(x)|dx& \ge \underset{M}{\overset{\mathrm{\infty }}{\int }}|f(x)|dx\\ & \ge \underset{M}{\overset{\mathrm{\infty }}{\int }}\frac{\delta }{2}dx\\ & =\mathrm{\infty }\end{array}$

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

$\underset{b\to \mathrm{\infty }}{\lim }f(b)=0$

Using the same reasoning as above,

$\underset{a\to -\mathrm{\infty }}{\lim }f(a)=0$

Hence,

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}^{\prime }(x)dx=\underset{b\to \mathrm{\infty }}{\lim }f(b)-\underset{a\to -\mathrm{\infty }}{\lim }f(a)=0$

Alternate Solution

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

$f(-m)+c-ϵ<f(m)<f(-m)+c+ϵ$ 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 $\underset{-\mathrm{\infty }}{\overset{-n}{\int }}+\underset{n}{\overset{\mathrm{\infty }}{\int }}f<\delta$. But by the above estimate,

$\underset{-\mathrm{\infty }}{\overset{-n}{\int }}f+\underset{n}{\overset{\mathrm{\infty }}{\int }}f>\underset{-\mathrm{\infty }}{\overset{-n}{\int }}f+\underset{-\mathrm{\infty }}{\overset{-n}{\int }}f+(c-ϵ)=\underset{-\mathrm{\infty }}{\overset{-n}{\int }}2f+\underset{-\mathrm{\infty }}{\overset{-n}{\int }}(c-ϵ)=\mathrm{\infty }$

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 $f_n(x)\to f(x)$ for almost every $x\in [0,1]$. Let $p>1$ and $M>0$ and suppose that $\Vert f_n{\Vert }_p\le M$ for all $n$ (a) Prove that $\Vert f{\Vert }_p\le M$. (b)Prove that $\Vert f-f_n{\Vert }_1\to 0$ as $n\to \mathrm{\infty }$

Solution 3a

By definition of norm,

$\Vert f{\Vert }_p={(\int |f(x){|}^p\mathrm{d}x)}^{\frac{1}{p}}$

Since $\Vert f_n{\Vert }_p\le M$,

$\Vert f_n{\Vert }_p^p\le M^p$

By Fatou's Lemma,

$\begin{array}{cc}\Vert f{\Vert }_p^p& =\underset{0}{\overset{1}{\int }}|f(x){|}^{p}dx\\ & =\underset{0}{\overset{1}{\int }}\underset{n}{\lim \inf }|f_n(x){|}^{p}dx\\ & \le \underset{n}{\lim \inf }\underset{0}{\overset{1}{\int }}|f_n(x){|}^{p}dx\\ & \le M^p\end{array}$

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

$\Vert f{\Vert }_p\le M$

Solution 3b

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

$\begin{array}{cc}\underset{A}{\int }|(f(x)-f_n(x))\cdot 1|dx& \le {(\underset{A}{\int }|f(x)-f_n(x){|}^{p}dx)}^{\frac{1}{p}}\cdot {\left(\underset{A}{\int }{1}^{q}dx\right)}^{\frac{1}{q}}\\ & \le {(\underset{A}{\int }|2f(x){|}^{p}dx)}^{\frac{1}{p}}\cdot {\left(\underset{A}{\int }{1}^{q}dx\right)}^{\frac{1}{q}}\\ & \le 2{(\underset{A}{\int }|f(x){|}^{p}dx)}^{\frac{1}{p}}\cdot {\left(\underset{A}{\int }{1}^{q}dx\right)}^{\frac{1}{q}}\\ & \le 2M\cdot (m(A){)}^{\frac{1}{q}}\end{array}$

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

Hence, $|f(x)-f_n(x)|\le 2M(m(A){)}^{\frac{1}{q}}$

The Vitali Convergence Theorem then implies

$\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{1}{\int }}|f(x)-f_n(x)|dx=\underset{0}{\overset{1}{\int }}\underset{n\to \mathrm{\infty }}{\lim }|f(x)-f_n(x)|dx=0$

Problem 5

Suppose $f(x),xf(x)\in L^2(R)$. Prove that $f(x)\in L^1(R)$ and that $\Vert f{\Vert }_1\le \sqrt{2}(\Vert f{\Vert }_2+\Vert xf{\Vert }_2)$

Solution 5

$\begin{array}{cc}\underset{R}{\int }|f(x)|dx& =\underset{|x|>1}{\int }|xf(x|\cdot \frac{1}{|x|}dx+\underset{|x|<1}{\int }|f(x)|\cdot 1dx\\ & \le {(\underset{|x|>1}{\int }|xf(x){|}^{2}dx)}^{\frac{1}{2}}\cdot {\left(\underset{|x|>1}{\int }\frac{1}{|x{|}^2}dx\right)}^{\frac{1}{2}}+{(\underset{|x|<1}{\int }|f(x){|}^{2}dx)}^{\frac{1}{2}}\cdot {\left(\underset{|x|<1}{\int }{1}^{2}dx\right)}^{\frac{1}{2}}\end{array}$