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

### Construct Contradiction

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,
$0=\lim _{M\rightarrow \infty }\int _{[-M,M]^{C}}|f|\geq {\frac {\epsilon _{0}}{2}}\delta (\epsilon _{0})\!\,$
which is a huge contradiction.
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
$\int _{x_{0}-\delta (\epsilon _{0})}^{x_{0}}|f|\!\,$
instead of the original one

## 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.

#### Proof
### 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

From the 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.