UMD Analysis Qualifying Exam/Jan08 Real

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.


 * $$ \lim_{|x| \rightarrow \infty} f(x) =0 \!\, $$

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 $$ x \in R \!\, $$

"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.

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.

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.