# UMD Analysis Qualifying Exam/Jan08 Real

## Problem 1

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

## Solution 1

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

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

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

or

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

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

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

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

Then, the negation of the above statement gives

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

### Apply Uniform Continuity

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

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

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

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

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

which implies

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

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

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

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

Then,

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

Therefore,

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

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

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

## Solution 1 (Alternate)

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

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

if

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

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

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

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

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

So we have ${\displaystyle \{I_{n}\}=\{(x_{n}-\delta (\epsilon _{0}),x_{n}+\delta (\epsilon _{0})\}\!\,}$  with ${\displaystyle I_{i}\cap I_{j}=\emptyset \!\,}$  if ${\displaystyle i\neq j\!\,}$  and ${\displaystyle |f(x)|\geq \epsilon _{0}\!\,}$  for all ${\displaystyle x\in I_{n}\!\,}$  and for all ${\displaystyle n\!\,}$ .

In other words, we are choosing disjoint subintervals of the real line that are of length ${\displaystyle 2\delta (\epsilon _{0})\!\,}$ , centered around each ${\displaystyle x_{i}\!\,}$  for ${\displaystyle i=1,2,3,\ldots \!\,}$ , and separated by at least ${\displaystyle \delta (\epsilon _{0})\!\,}$ .

Hence,

{\displaystyle {\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 ${\displaystyle f\in L^{1}(R)\!\,}$ .

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

${\displaystyle |f(x)|<\epsilon \!\,}$

i.e.

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

## Problem 3

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

## Solution 3

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

{\displaystyle {\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 ${\displaystyle f^{\prime }(t)=0\!\,}$  a.e.

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

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

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

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

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

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

${\displaystyle |c|=0\!\,}$

i.e. ${\displaystyle f(x)=0\!\,}$  for all ${\displaystyle x\in R\!\,}$

## Problem 5

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

## Problem 5a

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

## Solution 5a

### "One-half" triangle inequality

First, for all ${\displaystyle a,b\geq 0\!\,}$ ,

{\displaystyle {\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,

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

### L^1/2 is Linear Space

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

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

### L^1/2 is Metric Space

#### Non-negativity

${\displaystyle (i)\!\,}$  Since ${\displaystyle d(f,g)\geq 0\!\,}$ ,

#### Zero Distance

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

#### Triangle Inequality

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

{\displaystyle {\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 ${\displaystyle (i)\!\,}$  and ${\displaystyle (ii)\!\,}$  , we conclude that ${\displaystyle d(f,g)\!\,}$  is a metric space.

## Problem 5b

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

## Solution 5b

For ${\displaystyle a,b,c\geq 0\!\,}$ ,

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

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

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

### 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 ${\displaystyle \{f_{n}\}\!\,}$  such that for all ${\displaystyle n\!\,}$ ,

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

### Setup telescoping sum

Rewrite ${\displaystyle f_{m}(x)\!\,}$  as a telescoping sum (successive terms cancel out) i.e.

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

The triangle inequality implies,

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

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

### Define a sequence {g}_m

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

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

and

${\displaystyle g_{m}(x)\geq 0\!\,}$ .

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

Also,

{\displaystyle {\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 ${\displaystyle m\!\,}$

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

### Apply Monotone Convergence Theorem

By the Monotone Convergence Theorem,

{\displaystyle {\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,

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

### Apply Lebesgue Dominated Convergence Theorem

{\displaystyle {\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 ${\displaystyle f_{1},g_{m-1}\in L^{\frac {1}{2}}\!\,}$

Hence,

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

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