UMD Analysis Qualifying Exam/Jan10 Real

Problem 1

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

Solution 1

We will show that $\underset{n\to \mathrm{\infty }}{\lim }||f|{|}_{L^n}=||f|{|}_{L^{\mathrm{\infty }}}$

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,

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{\lim }(\int |f{|}^n{)}^{1/n}& =\underset{n\to \mathrm{\infty }}{\lim }(\int (\sum _{i=1}^N{a}_i{\mathrm{X}}_{E_i}{)}^n{)}^{1/n}\text{for notation, assume all }{a}_i\text{are positive.}\\ & =\underset{n\to \mathrm{\infty }}{\lim }(\int \sum _{i=1}^N{a}_i^n{\mathrm{X}}_{E_i}{)}^{1/n}\text{ since each of the cross terms equal zero since }{E}_i\text{ are all disjoint}\\ & =\underset{n\to \mathrm{\infty }}{\lim }(\sum _{i=1}^N\int a_i^n{\mathrm{X}}_{E_i}{)}^{1/n}\text{ by finite additivity of the Lebesgue integral}\\ & =\underset{n\to \mathrm{\infty }}{\lim }(\sum _{i=1}^N{a}_i^{n}m(E_i){)}^{1/n}\end{array}$

We now wish to compute this limit.

An upper bound is: $\sum _{i=1}^N{a}_i^{n}m(E_i){)}^{1/n}\le \sum _{i=1}^N(\underset{1\le j\le N}{\max }{a}_j^n)\cdot 1{)}^{1/n}$ since our function $f$ is defined only on the interval $[0,1]$. The right hand side, goes to $\max a_i$ as $n\to \mathrm{\infty }$.

A lower bound is: $\sum _{i=1}^N{a}_i^{n}m(E_i){)}^{1/n}\ge (\underset{1\le j\le N}{\max }{a}_j^n)m(E_i){)}^{1/n}$ since we assumed each of the $a_i$ and $m(E_i)$ to be positive. This also tends to $\max a_i$ as $n\to \mathrm{\infty }$.

Thus we have shown that $\underset{n\to \mathrm{\infty }}{\lim }(\int |f{|}^n{)}^{1/n}=\underset{1\le i\le N}{\max }{a}_i=||f|{|}_{L^{\mathrm{\infty }}([0,1])}$ for simple $f$. By density of the simple functions in $L^p$, this shows the same result for general $L^p$ functions.

So $\underset{n\to \mathrm{\infty }}{\lim }||f|{|}_{L^n}=||f|{|}_{L^{\mathrm{\infty }}}$. Now suppose $||f|{|}_{L^{\mathrm{\infty }}}=M>0$. Then for some $N$ we have $||f|{|}_{L^n}>M/2$ for every $n>N$. Thus

$\sum _{n=1}^{\mathrm{\infty }}||f|{|}_{L^n}\ge \sum _{n=N+1}^{\mathrm{\infty }}M/2=\mathrm{\infty }.$

Thus if we have any hope of $\sum _{n=1}^{\mathrm{\infty }}||f|{|}_{L^n}<\mathrm{\infty }$ we must have $||f|{|}_{L^{\mathrm{\infty }}}=0$ which implies that $f=0$ a.e. on [0,1].

Problem 3

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

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 L¹(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 L¹, 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 $L^1(ℝ)$ with $||f_n|{|}_1\le 1$ for all $n$ and $\underset{n\to \mathrm{\infty }}{\lim }{f}_n(x)=f(x)a.e.$ (a) Prove that $f\in L^1(ℝ)$ and that $||f|{|}_1\le 1$ (b) Show that $\underset{n\to \mathrm{\infty }}{\lim }(||f-f_n|{|}_1-||f_n|{|}_1+||f_n|{|}_1)=0$.