# UMD Analysis Qualifying Exam/Jan10 Real
## Problem 1

* Assume that $n\geq 1$ is an integer, and let $f\in \cap _{n=1}^{\infty }L^{n}([0,1])$. Prove that if $\sum _{n=1}^{\infty }||f||_{L^{n}([0,1])}<\infty$, then $f=0$ a.e.

### Solution 1

We will show that $\lim _{n\to \infty }||f||_{L^{n}}=||f||_{L^{\infty }}$
Since the simple functions are dense in $L^{p}$, it suffices to show the result where $f=\sum _{i=1}^{N}a_{i}\mathrm {X} _{E_{i}}$. Without loss of generality, we can assume that $\{E_{i}\}$ is a disjoint family of measurable sets. Then,
${\begin{aligned}\lim _{n\to \infty }(\int |f|^{n})^{1/n}&=\lim _{n\to \infty }(\int (\sum _{i=1}^{N}a_{i}\mathrm {X} _{E_{i}})^{n})^{1/n}{\text{for notation, assume all }}a_{i}{\text{are positive.}}\\&=\lim _{n\to \infty }(\int \sum _{i=1}^{N}a_{i}^{n}\mathrm {X} _{E_{i}})^{1/n}{\text{ since each of the cross terms equal zero since }}E_{i}{\text{ are all disjoint}}\\&=\lim _{n\to \infty }(\sum _{i=1}^{N}\int a_{i}^{n}\mathrm {X} _{E_{i}})^{1/n}{\text{ by finite additivity of the Lebesgue integral}}\\&=\lim _{n\to \infty }(\sum _{i=1}^{N}a_{i}^{n}m(E_{i}))^{1/n}\end{aligned}}$
We now wish to compute this limit.
An upper bound is: $\sum _{i=1}^{N}a_{i}^{n}m(E_{i}))^{1/n}\leq \sum _{i=1}^{N}(\max _{1\leq j\leq N}a_{j}^{n})\cdot 1)^{1/n}$ since our function $f$ is defined only on the interval $[0,1]$. The right hand side, goes to $\max a_{i}$ as $n\to \infty$.
A lower bound is: $\sum _{i=1}^{N}a_{i}^{n}m(E_{i}))^{1/n}\geq (\max _{1\leq j\leq N}a_{j}^{n})m(E_{i}))^{1/n}$ since we assumed each of the $a_{i}$ and $m(E_{i})$ to be positive. This also tends to $\max a_{i}$ as $n\to \infty$.
Thus we have shown that $\lim _{n\to \infty }(\int |f|^{n})^{1/n}=\max _{1\leq i\leq N}a_{i}=||f||_{L^{\infty }([0,1])}$ for simple $f$. By density of the simple functions in $L^{p}$, this shows the same result for general $L^{p}$ functions.
So $\lim _{n\to \infty }||f||_{L^{n}}=||f||_{L^{\infty }}$. Now suppose $||f||_{L^{\infty }}=M>0$. Then for some $N$ we have $||f||_{L^{n}}>M/2$ for every $n>N$. Thus
$\sum _{n=1}^{\infty }||f||_{L^{n}}\geq \sum _{n=N+1}^{\infty }M/2=\infty .$
Thus if we have any hope of $\sum _{n=1}^{\infty }||f||_{L^{n}}<\infty$ we must have $||f||_{L^{\infty }}=0$ which implies that $f=0$ a.e. on [0,1].

## Problem 3

* Let $f\in L^{1}(\mathbb {R} )$.
* (a) Determine
* $\lim _{x\to 0}\int _{\mathbb {R} }|f(t+x)+f(t)|\,dt$
* (b) Determine
* $\lim _{x\to \infty }\int _{\mathbb {R} }|f(t+x)+f(t)|\,dt$

### Solution 3

For both parts (a) and (b), we have a clear upper bound of 2||f||$_1$, by the triangle inequality. It remains to be shown that this is a tight upper bound in both cases.
The same approach can be used for both, with minor changes for the two different limits. Since step functions are dense in L$^1$(R), pick epsilon e>0 and approximate f by some step function g, such that ||f-g||$_1$<e. Let f$_x$(t)=f(x+t), g$_x$(t)=g(t+x).
Then ||f$_x$+f|| = ||f$_x$+f+(g$_x$+g)-(g$_x$+g)||. By the triangle inequality, this is greater than or equal to ||g$_x$+g||-||f$_x$+f-(g$_x$+g)||. By yet another application of the triangle inequality, the second term is greater than or equal to -2e. The proof diverges at this point.
For part (a), for any particular t, x, |g$_x$(t)+g(t)| is less than |g$_x$(t)|+|g(t)| if and only if g$_x$(t) and g(t) have opposite signs. Since g is a step function, this can clearly only happen when t and t+x are in intervals with different coefficients, and hence can happen at most for a distance of x per interval, across a finite number n of intervals. Since the difference for any particular step function g is bounded by M=max(g)-min(g), we get the following inequality:
||g$_x$+g|| >= ||g$_x$||+||g|| - x*n*M. Clearly, the limit of this as x goes to 0 is 2||g||, which is itself bounded below by 2||f||-2e. Adding up the two parts, we get a lower bound of the limit: lim$_{x->0}$||f$_x$+f||>=2||f||-4e, for any positive epsilon. Thus the bound of 2||f|| is tight.
The justification in part (b) is simpler. Since g is a step function in L$^1$, it has bounded support, so some value of x will be large enough that g$_x$ and g have disjoint supports. Hence we can just say that the limit is ||g$_x$||+||g||>=2||f||-2e.

## Problem 5

* Suppose that $\{f_{n}\}$ is a sequence in $L^{1}(\mathbb {R} )$ with $||f_{n}||_{1}\leq 1$ for all $n$ and
* $\lim _{n\to \infty }f_{n}(x)=f(x)\,a.e.$
* (a) Prove that $f\in L^{1}(\mathbb {R} )$ and that $||f||_{1}\leq 1$
* (b) Show that
* $\lim _{n\to \infty }(||f-f_{n}||_{1}-||f_{n}||_{1}+||f_{n}||_{1})=0$.