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.

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.