UMD Analysis Qualifying Exam/Jan08 Real

Problem 1

Suppose that $f\in L^1(R)$ is a uniformly continous function. Show that $\underset{|x|\to \mathrm{\infty }}{\lim }f(x)=0$

Solution 1

L^1 implies integral of tail end of function goes to zero

$\begin{array}{cc}\underset{M\to \mathrm{\infty }}{\lim }\underset{[-M,M]}{\int }|f|& =\underset{ℝ}{\int }|f|\\ \underset{M\to \mathrm{\infty }}{\lim }\underset{[-M,M]}{\int }|f|-\underset{<\mathrm{\infty }\text{ since }f\in L^1}{\underbrace{\underset{ℝ}{\int }|f|}}& =0\\ \underset{ℝ}{\int }|f|-\underset{M\to \mathrm{\infty }}{\lim }\underset{[-M,M]}{\int }|f|& =0\\ \underset{M\to \mathrm{\infty }}{\lim }\underset{[-M,M{]}^c}{\int }|f|& =0\end{array}$

Assume Not

Suppose $\underset{|x|\to \mathrm{\infty }}{\lim }f(x)\ne 0$. Then,

$\underset{x\to \mathrm{\infty }}{\lim }f(x)\ne 0$

or

$\underset{x\to -\mathrm{\infty }}{\lim }f(x)\ne 0$

Without loss of generality, we can assume the first one, i.e., $\underset{x\to \mathrm{\infty }}{\lim }f(x)\ne 0$ (see remark below to see why this)

Note that $\underset{x\to \mathrm{\infty }}{\lim }f(x)=0$ can be written as

$\mathrm{\forall }ϵ>0[\mathrm{\exists }M>0[\mathrm{\forall }x>M[|f(x)|<ϵ]]]$

Then, the negation of the above statement gives

$\mathrm{\exists }{ϵ}_0>0[\mathrm{\forall }M>0[\mathrm{\exists }{x}_0>M[|f(x_0)|>{ϵ}_0]]]$

Apply Uniform Continuity

Because of the uniform continuity, for the ${ϵ}_0$ there is a $\delta ({ϵ}_0)>0$ such that

$|f(x_0)-f(x)|<\frac{{ϵ}_0}{2}$ ,

whenever $|x-x_0|<\delta ({ϵ}_0)$

Then, if $|x-x_0|<\delta ({ϵ}_0)$, by Triangle Inequality, we have

${ϵ}_0<|f(x_0)|<|f(x)|+|f(x_0)-f(x)|<|f(x)|+\frac{{ϵ}_0}{2}$

which implies

$\frac{{ϵ}_0}{2}<|f(x)|$,

whenever $|x-x_0|<\delta ({ϵ}_0)$

Construct Contradiction

Let $x_0$ be a number greater than $M$. Note that ${ϵ}_0$ and $\delta ({ϵ}_0)$ do not depend on $M$. With this in mind, note that

$\underset{[-M,M{]}^C}{\int }|f|>\underset{x_0}{\overset{x_0+\delta ({ϵ}_0)}{\int }}|f|>\frac{{ϵ}_0}{2}\delta ({ϵ}_0)\text{(*)}$

Then,

$0=\underset{M\to \mathrm{\infty }}{\lim }\underset{[-M,M{]}^C}{\int }|f|\ge \frac{{ϵ}_0}{2}\delta ({ϵ}_0)$

which is a huge contradiction.

Therefore,

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

Remark If we choose to work with the assumption that $\underset{x\to -\mathrm{\infty }}{\lim }|f|\ne 0$ , then in (*), we just need to work with

$\underset{x_0-\delta ({ϵ}_0)}{\overset{x_0}{\int }}|f|$

instead of the original one

Solution 1 (Alternate)

By uniform continuity, for all $ϵ>0$, there exists $\delta (ϵ)>0$ such that for all $x_1,x_2\in R$,

$|f(x_1)-f(x_2)|<ϵ$

if

$|x_1-x_2|<\delta (ϵ)$

Assume for the sake of contradiction there exists ${ϵ}_0>0$ such that for all $M>0$, there exists $x\in R$ such that $|x|>M$ and $|f(x)|\ge {ϵ}_0$.

Let $M=1$, then there exists $x_1$ such that $|x_1|>M$ and $|f(x_1)|\ge {ϵ}_0$.

Let $M=|x_1|+3\delta ({ϵ}_0)$, then there exists $x_2$ such that $|x_2|>M$ and $|f(x_2)|\ge {ϵ}_0$.

Let $M=|x_n|+3\delta ({ϵ}_0)$, then there exists $x_{n+1}$ such that $|x_{n+1}|>M$ and $|f(x_{n+2})|\ge {ϵ}_0$.

So we have $\{I_n\}=\{(x_n-\delta ({ϵ}_0),x_n+\delta ({ϵ}_0)\}$ with $I_i\cap I_j=\mathrm{\varnothing }$ if $i\ne j$ and $|f(x)|\ge {ϵ}_0$ for all $x\in I_n$ and for all $n$.

In other words, we are choosing disjoint subintervals of the real line that are of length $2\delta ({ϵ}_0)$, centered around each $x_i$ for $i=1,2,3,\dots$, and separated by at least $\delta ({ϵ}_0)$.

Hence,

$\begin{array}{cc}\underset{R}{\int }|f(x)|dx& \ge \sum _{n=1}^{\mathrm{\infty }}\underset{I_n}{\int }|f(x)|dx\\ & \ge \sum _{n=1}^{\mathrm{\infty }}{ϵ}_0\underset{I_n}{\int }dx\\ & =\sum _{n=1}^{\mathrm{\infty }}{ϵ}_0\cdot 2\delta ({ϵ}_0)\\ & =+\mathrm{\infty }\end{array}$

which contradicts the assumption that $f\in L^1(R)$.

Therefore, for all $ϵ>0$ there exists $M>0$ such that for all $|x|>M$,

$|f(x)|<ϵ$

i.e.

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

Problem 3

Suppose $f$ is absolutely continuous on $R$, and $f\in L^1(R)$. Show that if in addition $\underset{t\to 0^{+}}{\lim }\underset{R}{\int }\left|\frac{f(x+t)-f(x)}{t}\right|dx=0$ then $f=0$

Solution 3

By absolute continuity, Fatou's Lemma, and hypothesis we have

$\begin{array}{cc}\underset{R}{\int }|f^{\prime }(t)|dt& =\underset{R}{\int }\underset{t\to 0^{+}}{\lim \inf }\left|\frac{f(x+t)-f(x)}{t}\right|dx\\ & \le \underset{t\to 0^{+}}{\lim \inf }\underset{R}{\int }\left|\frac{f(x+t)-f(x)}{t}\right|dx\\ & =0\end{array}$

Hence $f^{\prime }(t)=0$ a.e.

From the fundamental theorem of calculus, for all $x\in R$,

$f(x)-f(0)=\underset{0}{\overset{x}{\int }}{f}^{\prime }(t)dt=0$

i.e. $f(x)$ is a constant $c=f(0)$.

Assume for the sake of contradiction that $|c|>0$, then

$\underset{R}{\int }|f(x)|dx=\underset{R}{\int }|c|dx=\mathrm{\infty }$.

which contradicts the hypothesis $f\in L^1(R)$. Hence,

$|c|=0$

i.e. $f(x)=0$ for all $x\in R$

Problem 5

Suppose that $L^{\frac{1}{2}}(R)$ is the set of all equivalence classes of measurable functions for which $\underset{R}{\int }|f(x){|}^{\frac{1}{2}}dx<\mathrm{\infty }$

Problem 5a

Show that it is a metric linear space with the metric $d(f,g)=\underset{R}{\int }|f(x)-g(x){|}^{\frac{1}{2}}dx$ where $f,g\in L^{\frac{1}{2}}(R)$.

Solution 5a

"One-half" triangle inequality

First, for all $a,b\ge 0$,

$\begin{array}{cc}(a+b)& \le (a+b)+2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\\ & =a+2a^{\frac{1}{2}}{b}^{\frac{1}{2}}+b\\ & =(a^{\frac{1}{2}}{)}^2+2a^{\frac{1}{2}}{b}^{\frac{1}{2}}+(b^{\frac{1}{2}}{)}^2\\ & =(a^{\frac{1}{2}}+b^{\frac{1}{2}}{)}^2\end{array}$

Taking square roots of both sides of the inequality yields,

$(a+b{)}^{\frac{1}{2}}\le a^{\frac{1}{2}}+b^{\frac{1}{2}}$

L^1/2 is Linear Space

Hence for all $f,g\in L^{\frac{1}{2}}$,

$\begin{array}{cc}\underset{R}{\int }|af(x)+bg(x){|}^{\frac{1}{2}}dx& \le \underset{R}{\int }(|a{|}^{\frac{1}{2}}|f(x){|}^{\frac{1}{2}}+|b{|}^{\frac{1}{2}}|g(x){|}^{\frac{1}{2}})dx\\ & =|a{|}^{\frac{1}{2}}\underset{R}{\int }|f(x){|}^{\frac{1}{2}}dx+|b{|}^{\frac{1}{2}}\underset{R}{\int }|g(x){|}^{\frac{1}{2}}dx\\ & <\mathrm{\infty }\end{array}$

Hence, $L^{\frac{1}{2}}$ is a linear space.

L^1/2 is Metric Space

Non-negativity

$(i)$ Since $d(f,g)\ge 0$,

Zero Distance

$d(f,g)=0⟺f=ga.e.$

Triangle Inequality

$(ii)$ Also, for all $f,g,h\in L^{\frac{1}{2}}$,

$\begin{array}{cc}d(f,g)& =\underset{R}{\int }|f(x)-g(x){|}^{\frac{1}{2}}dx\\ & =\underset{R}{\int }(|f(x)-h(x)+h(x)-g(x){|}^{\frac{1}{2}}dx\\ & \le \underset{R}{\int }(|f(x)-h(x)|+|h(x)-g(x)|{)}^{\frac{1}{2}}dx\\ & \le \underset{R}{\int }|f(x)-h(x){|}^{\frac{1}{2}}dx+\underset{R}{\int }|h(x)-g(x){|}^{\frac{1}{2}}dx\\ & =d(f,h)+d(h,g)\end{array}$

From $(i)$ and $(ii)$ , we conclude that $d(f,g)$ is a metric space.

Problem 5b

Show that with this metric $L^{\frac{1}{2}}(R)$ is complete.

Solution 5b

For $a,b,c\ge 0$,

$(a+b+c{)}^{\frac{1}{2}}\le a^{\frac{1}{2}}+(b+c{)}^{\frac{1}{2}}\le a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}$

By induction, we then have for all $a_k\ge 0$ and all $k$

${\left(\sum _{k=1}^n{a}_k\right)}^{\frac{1}{2}}\le \sum _{k=1}^n{a}_k^{\frac{1}{2}}$

Work with Subsequence of Cauchy Sequence

We can equivalently prove completeness by showing that a subsequence of a Cauchy sequence converges.

Claim

If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges.

Construct a subsequence

Choose $\{f_n\}$ such that for all $n$,

$d(f_n,f_{n+1})<\frac{1}{2^n}$

Setup telescoping sum

Rewrite $f_m(x)$ as a telescoping sum (successive terms cancel out) i.e.

$f_m(x)=f_1(x)+\underset{g_{m-1}(x)}{\underbrace{\sum _{n=1}^{m-1}(f_{n+1}(x)-f_n(x))}}$.

The triangle inequality implies,

$|f_m(x){|}^{\frac{1}{2}}\le |f_1(x){|}^{\frac{1}{2}}+|g_{m-1}(x){|}^{\frac{1}{2}}$

which means the sequence $|f_m(x){|}^{\frac{1}{2}}$ is always dominated by the sequence on the right hand side of the inequality.

Define a sequence {g}_m

Let $g_m(x)=\sum _{n=1}^m|f_{n+1}(x)-f_n(x)|$, then

$g_m(x)\le g_{m+1}(x)$

and

$g_m(x)\ge 0$.

In other words, $\{g_m\}$ is a sequence of increasing, non-negative functions. Note that $g$, the limit of $\{g_m\}$ as $m\to \mathrm{\infty }$, exists since $\{g_m\}$ is increasing. ($g$ is either a finite number $L$ or $\mathrm{\infty }$.)

Also,

$\begin{array}{cc}\underset{R}{\int }|g_m(x){|}^{\frac{1}{2}}dx& =\underset{R}{\int }\sum _{n=1}^m|f_{n+1}(x)-f_n(x){|}^{\frac{1}{2}}dx\\ & =\sum _{n=1}^m\underset{d(f_{n+1},f_n)}{\underbrace{\underset{R}{\int }|f_{n+1}(x)-f_n(x){|}^{\frac{1}{2}}dx}}\\ & \le \sum _{n=1}^m\frac{1}{2^n}\\ & \le 1\end{array}$

Hence, for all $m$

$\underset{R}{\int }|g_m(x){|}^{\frac{1}{2}}dx\le 1$

Apply Monotone Convergence Theorem

By the Monotone Convergence Theorem,

$\begin{array}{cc}\underset{R}{\int }\underset{m\to \mathrm{\infty }}{\lim }|g_m(x){|}^{\frac{1}{2}}dx& =\underset{m\to \mathrm{\infty }}{\lim }\underset{R}{\int }|g_m(x){|}^{\frac{1}{2}}dx\\ & \le 1\\ & <\mathrm{\infty }\end{array}$

Hence,

$\underset{m\to \mathrm{\infty }}{\lim }{g}_m(x)\in L^{\frac{1}{2}}$

Apply Lebesgue Dominated Convergence Theorem

From the Lebesgue dominated convergence theorem,

$\begin{array}{cc}\underset{R}{\int }\underset{m\to \mathrm{\infty }}{\lim }|f_m{|}^{\frac{1}{2}}& =\underset{m\to \mathrm{\infty }}{\lim }\underset{R}{\int }|f_m{|}^{\frac{1}{2}}\\ & \le \underset{m\to \mathrm{\infty }}{\lim }\underset{R}{\int }|f_1(x){|}^{\frac{1}{2}}+\underset{R}{\int }|g_{m-1}(x){|}^{\frac{1}{2}}\\ & <\mathrm{\infty }\end{array}$

where the last step follows since $f_1,g_{m-1}\in L^{\frac{1}{2}}$

Hence,

$\underset{m\to \mathrm{\infty }}{\lim }{f}_m\in L^{\frac{1}{2}}$

i.e. $L^{\frac{1}{2}}$ is complete.