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

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

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,

$\begin{align} \int_R |f(x)|dx &\geq \sum_{n=1}^{\infty} \int_{I_n} |f(x)|dx \\ &\geq \sum_{n=1}^{\infty} \epsilon_0 \int_{I_n} dx \\ &= \sum_{n=1}^{\infty} \epsilon_0 \cdot 2 \delta(\epsilon_0) \\ &= +\infty \end{align}$

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

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

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 $a,b \geq 0 \!\,$,

$\begin{align} (a+b) &\leq (a+b) + 2 a^{\frac{1}{2}} b^{\frac{1}{2}} \\ &= a+ 2 a^{\frac{1}{2}} b^{\frac{1}{2}}+b \\ &= (a^{\frac{1}{2}})^2+ 2 a^{\frac{1}{2}} b^{\frac{1}{2}}+(b^{\frac{1}{2}})^2 \\ &= (a^{\frac{1}{2}}+b^{\frac{1}{2}})^2 \end{align}$

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

$\begin{align} \int_R |af(x)+bg(x)|^{\frac{1}{2}} dx &\leq \int_R ( |a|^{\frac{1}{2}}|f(x)|^{\frac{1}{2}}+|b|^{\frac{1}{2}}|g(x)|^{\frac{1}{2}})dx \\ &=|a|^{\frac{1}{2}}\int_R |f(x)|^{\frac{1}{2}}dx+|b|^{\frac{1}{2}}\int_R |g(x)|^{\frac{1}{2}}dx \\ &< \infty \end{align}$

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 $f,g,h \in L^{\frac{1}{2}}\!\,$,

$\begin{align} d(f,g) &= \int_R |f(x)-g(x)|^{\frac{1}{2}} dx \\ &= \int_R (|f(x)-h(x)+h(x)-g(x)|^{\frac{1}{2}}dx \\ &\leq \int_R (|f(x)-h(x)|+|h(x)-g(x)|)^{\frac{1}{2}}dx \\ &\leq \int_R |f(x)-h(x)|^{\frac{1}{2}}dx + \int_R |h(x)-g(x)|^{\frac{1}{2}}dx \\ &=d(f,h)+d(h,g) \end{align}$

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 $|f_m(x)|^\frac12 \!\,$ 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,

$\begin{align} \int_R |g_m(x)|^{\frac{1}{2}}dx &= \int_R \sum_{n=1}^m |f_{n+1}(x)-f_n(x)|^{\frac{1}{2}} dx \\ &= \sum_{n=1}^m \underbrace{\int_R |f_{n+1}(x)-f_n(x)|^{\frac{1}{2}} dx }_{d(f_{n+1},f_n)}\\ &\leq \sum_{n=1}^m \frac{1}{2^n} \\ &\leq 1 \end{align}$

Hence, for all $m \!\,$

$\int_R |g_m(x)|^{\frac{1}{2}}dx \leq 1 \!\,$

Apply Monotone Convergence Theorem

By the Monotone Convergence Theorem,

$\begin{align} \int_R \lim_{m \rightarrow \infty} |g_m(x)|^{\frac{1}{2}}dx &= \lim_{m \rightarrow \infty} \int_R |g_m(x)|^{\frac{1}{2}}dx \\ &\leq 1 \\ &< \infty \end{align}$

Hence,

$\lim_{m \rightarrow \infty} g_m(x) \in L^{\frac{1}{2}} \!\,$

Apply Lebesgue Dominated Convergence Theorem

From the Lebesgue dominated convergence theorem,

$\begin{align} \int_R \lim_{m \rightarrow \infty} |f_m|^\frac12 &= \lim_{m \rightarrow \infty} \int_R |f_m|^{\frac12} \\ &\leq \lim_{m \rightarrow \infty} \int_R |f_1(x)|^{\frac12}+\int_R|g_{m-1}(x)|^{\frac12} \\ &< \infty \end{align}$

where the last step follows since $f_1, g_{m-1} \in L^{\frac12} \!\,$

Hence,

$\lim_{m \rightarrow \infty} f_m \in L^{\frac{1}{2}} \!\,$

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