# UMD Analysis Qualifying Exam/Jan08 Real

## Problem 1

 Suppose that $f\in L^{1}(R)\!\,$ is a uniformly continous function. Show that $\lim _{|x|\rightarrow \infty }f(x)=0\!\,$ ## Solution 1

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

{\begin{aligned}\lim _{M\rightarrow \infty }\int _{[-M,M]}|f|&=\int _{\mathbb {R} }|f|\\\lim _{M\rightarrow \infty }\int _{[-M,M]}|f|-\underbrace {\int _{\mathbb {R} }|f|} _{<\infty {\mbox{ since }}f\in L^{1}}&=0\\\int _{\mathbb {R} }|f|-\lim _{M\rightarrow \infty }\int _{[-M,M]}|f|&=0\\\lim _{M\rightarrow \infty }\int _{[-M,M]^{c}}|f|&=0\end{aligned}}\!\,

### Assume Not

Suppose $\lim _{|x|\rightarrow \infty }f(x)\neq 0\!\,$ . Then,

$\lim _{x\rightarrow \infty }f(x)\neq 0\!\,$

or

$\lim _{x\rightarrow -\infty }f(x)\neq 0\!\,$

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

Note that $\lim _{x\rightarrow \infty }f(x)=0\!\,$  can be written as

$\forall \epsilon >0\left[\exists M>0\left[\forall x>M\left[|f(x)|<\epsilon \right]\right]\right]\!\,$

Then, the negation of the above statement gives

$\exists \epsilon _{0}>0[\forall M>0[\exists x_{0}>M[|f(x_{0})|>\epsilon _{0}]]]\!\,$

### Apply Uniform Continuity

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

$|f(x_{0})-f(x)|<{\frac {\epsilon _{0}}{2}}\!\,$  ,

whenever $|x-x_{0}|<\delta (\epsilon _{0})\!\,$

Then, if $|x-x_{0}|<\delta (\epsilon _{0})\!\,$ , by Triangle Inequality, we have

$\epsilon _{0}<|f(x_{0})|<|f(x)|+|f(x_{0})-f(x)|<|f(x)|+{\frac {\epsilon _{0}}{2}}\!\,$

which implies

${\frac {\epsilon _{0}}{2}}<|f(x)|\!\,$ ,

whenever $|x-x_{0}|<\delta (\epsilon _{0})\!\,$

Let $x_{0}\!\,$  be a number greater than $M\!\,$ . Note that $\epsilon _{0}\!\,$  and $\delta (\epsilon _{0})\!\,$  do not depend on $M\!\,$ . With this in mind, note that

$\int _{[-M,M]^{C}}|f|>\int _{x_{0}}^{x_{0}+\delta (\epsilon _{0})}|f|>{\frac {\epsilon _{0}}{2}}\delta (\epsilon _{0})\qquad {\mbox{(*)}}\!\,$

Then,

$0=\lim _{M\rightarrow \infty }\int _{[-M,M]^{C}}|f|\geq {\frac {\epsilon _{0}}{2}}\delta (\epsilon _{0})\!\,$

Therefore,

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

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

$\int _{x_{0}-\delta (\epsilon _{0})}^{x_{0}}|f|\!\,$

## Solution 1 (Alternate)

By uniform continuity, for all $\epsilon >0\!\,$ , there exists $\delta (\epsilon )>0\!\,$  such that for all $x_{1},x_{2}\in R\!\,$ ,

$|f(x_{1})-f(x_{2})|<\epsilon \!\,$

if

$|x_{1}-x_{2}|<\delta (\epsilon )\!\,$

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

Let $M=1\!\,$ , then there exists $x_{1}\!\,$  such that $|x_{1}|>M\!\,$  and $|f(x_{1})|\geq \epsilon _{0}\!\,$ .

Let $M=|x_{1}|+3\delta (\epsilon _{0})\!\,$ , then there exists $x_{2}\!\,$  such that $|x_{2}|>M\!\,$  and $|f(x_{2})|\geq \epsilon _{0}\!\,$ .

Let $M=|x_{n}|+3\delta (\epsilon _{0})\!\,$ , then there exists $x_{n+1}\!\,$  such that $|x_{n+1}|>M\!\,$  and $|f(x_{n+2})|\geq \epsilon _{0}\!\,$ .

So we have $\{I_{n}\}=\{(x_{n}-\delta (\epsilon _{0}),x_{n}+\delta (\epsilon _{0})\}\!\,$  with $I_{i}\cap I_{j}=\emptyset \!\,$  if $i\neq j\!\,$  and $|f(x)|\geq \epsilon _{0}\!\,$  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 $2\delta (\epsilon _{0})\!\,$ , centered around each $x_{i}\!\,$  for $i=1,2,3,\ldots \!\,$ , and separated by at least $\delta (\epsilon _{0})\!\,$ .

Hence,

{\begin{aligned}\int _{R}|f(x)|dx&\geq \sum _{n=1}^{\infty }\int _{I_{n}}|f(x)|dx\\&\geq \sum _{n=1}^{\infty }\epsilon _{0}\int _{I_{n}}dx\\&=\sum _{n=1}^{\infty }\epsilon _{0}\cdot 2\delta (\epsilon _{0})\\&=+\infty \end{aligned}}

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

Therefore, for all $\epsilon >0\!\,$  there exists $M>0\!\,$  such that for all $|x|>M\!\,$ ,

$|f(x)|<\epsilon \!\,$

i.e.

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

## Problem 3

 Suppose $f\!\,$ is absolutely continuous on $R\!\,$ , and $f\in L^{1}(R)\!\,$ . Show that if in addition $\lim _{t\rightarrow 0^{+}}\int _{R}\left|{\frac {f(x+t)-f(x)}{t}}\right|dx=0\!\,$ then $f=0\!\,$ ## Solution 3

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

{\begin{aligned}\int _{R}|f^{\prime }(t)|dt&=\int _{R}{\underset {t\rightarrow 0^{+}}{\lim \inf }}\left|{\frac {f(x+t)-f(x)}{t}}\right|dx\\&\leq {\underset {t\rightarrow 0^{+}}{\lim \inf }}\int _{R}\left|{\frac {f(x+t)-f(x)}{t}}\right|dx\\&=0\end{aligned}}

Hence $f^{\prime }(t)=0\!\,$  a.e.

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

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

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

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

$\int _{R}|f(x)|dx=\int _{R}|c|dx=\infty \!\,$ .

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 $\int _{R}|f(x)|^{\frac {1}{2}}dx<\infty \!\,$ ## Problem 5a

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

## Solution 5a

### "One-half" triangle inequality

First, for all $a,b\geq 0\!\,$ ,

{\begin{aligned}(a+b)&\leq (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{aligned}}

Taking square roots of both sides of the inequality yields,

$(a+b)^{\frac {1}{2}}\leq a^{\frac {1}{2}}+b^{\frac {1}{2}}\!\,$

### L^1/2 is Linear Space

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

{\begin{aligned}\int _{R}|af(x)+bg(x)|^{\frac {1}{2}}dx&\leq \int _{R}(|a|^{\frac {1}{2}}|f(x)|^{\frac {1}{2}}+|b|^{\frac {1}{2}}|g(x)|^{\frac {1}{2}})dx\\&=|a|^{\frac {1}{2}}\int _{R}|f(x)|^{\frac {1}{2}}dx+|b|^{\frac {1}{2}}\int _{R}|g(x)|^{\frac {1}{2}}dx\\&<\infty \end{aligned}}

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

### L^1/2 is Metric Space

#### Non-negativity

$(i)\!\,$  Since $d(f,g)\geq 0\!\,$ ,

#### Zero Distance

$d(f,g)=0\quad \Longleftrightarrow \quad f=g\,\,a.e.\!\,$

#### Triangle Inequality

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

{\begin{aligned}d(f,g)&=\int _{R}|f(x)-g(x)|^{\frac {1}{2}}dx\\&=\int _{R}(|f(x)-h(x)+h(x)-g(x)|^{\frac {1}{2}}dx\\&\leq \int _{R}(|f(x)-h(x)|+|h(x)-g(x)|)^{\frac {1}{2}}dx\\&\leq \int _{R}|f(x)-h(x)|^{\frac {1}{2}}dx+\int _{R}|h(x)-g(x)|^{\frac {1}{2}}dx\\&=d(f,h)+d(h,g)\end{aligned}}

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 $a,b,c\geq 0\!\,$ ,

$(a+b+c)^{\frac {1}{2}}\leq a^{\frac {1}{2}}+(b+c)^{\frac {1}{2}}\leq a^{\frac {1}{2}}+b^{\frac {1}{2}}+c^{\frac {1}{2}}\!\,$

By induction, we then have for all $a_{k}\geq 0\!\,$  and all $k\!\,$

$\left(\sum _{k=1}^{n}a_{k}\right)^{\frac {1}{2}}\leq \sum _{k=1}^{n}a_{k}^{\frac {1}{2}}\!$

### 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.

$f_{m}(x)=f_{1}(x)+\underbrace {\sum _{n=1}^{m-1}(f_{n+1}(x)-f_{n}(x))} _{g_{m-1}(x)}\!\,$ .

The triangle inequality implies,

$|f_{m}(x)|^{\frac {1}{2}}\leq |f_{1}(x)|^{\frac {1}{2}}+|g_{m-1}(x)|^{\frac {1}{2}}\!\,$

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

$g_{m}(x)\leq g_{m+1}(x)\!\,$

and

$g_{m}(x)\geq 0\!\,$ .

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

Also,

{\begin{aligned}\int _{R}|g_{m}(x)|^{\frac {1}{2}}dx&=\int _{R}\sum _{n=1}^{m}|f_{n+1}(x)-f_{n}(x)|^{\frac {1}{2}}dx\\&=\sum _{n=1}^{m}\underbrace {\int _{R}|f_{n+1}(x)-f_{n}(x)|^{\frac {1}{2}}dx} _{d(f_{n+1},f_{n})}\\&\leq \sum _{n=1}^{m}{\frac {1}{2^{n}}}\\&\leq 1\end{aligned}}

Hence, for all $m\!\,$

$\int _{R}|g_{m}(x)|^{\frac {1}{2}}dx\leq 1\!\,$

### Apply Monotone Convergence Theorem

By the Monotone Convergence Theorem,

{\begin{aligned}\int _{R}\lim _{m\rightarrow \infty }|g_{m}(x)|^{\frac {1}{2}}dx&=\lim _{m\rightarrow \infty }\int _{R}|g_{m}(x)|^{\frac {1}{2}}dx\\&\leq 1\\&<\infty \end{aligned}}

Hence,

$\lim _{m\rightarrow \infty }g_{m}(x)\in L^{\frac {1}{2}}\!\,$

### Apply Lebesgue Dominated Convergence Theorem

{\begin{aligned}\int _{R}\lim _{m\rightarrow \infty }|f_{m}|^{\frac {1}{2}}&=\lim _{m\rightarrow \infty }\int _{R}|f_{m}|^{\frac {1}{2}}\\&\leq \lim _{m\rightarrow \infty }\int _{R}|f_{1}(x)|^{\frac {1}{2}}+\int _{R}|g_{m-1}(x)|^{\frac {1}{2}}\\&<\infty \end{aligned}}

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

Hence,

$\lim _{m\rightarrow \infty }f_{m}\in L^{\frac {1}{2}}\!\,$

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