UMD Analysis Qualifying Exam/Jan10 Real

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 $$f=\sum_{i=1}^N a_i \Chi_{E_i}$$. Without loss of generality, we can assume that $$\{E_i\}$$ is a disjoint family of measurable sets. Then,

$$ \begin{align} \lim_{n\to\infty} (\int |f|^n)^{1/n}&=\lim_{n\to\infty} (\int (\sum_{i=1}^N a_i \Chi_{E_i})^n)^{1/n}\text{for notation, assume all } a_i \text{are positive.}\\ &=\lim_{n\to\infty} (\int \sum_{i=1}^N a_i^n \Chi_{E_i})^{1/n} \text{ since each of the cross terms equal zero since } E_i \text{ are all disjoint}\\ &=\lim_{n\to\infty} (\sum_{i=1}^N \int a_i^n \Chi_{E_i})^{1/n}\text{ by finite additivity of the Lebesgue integral}\\ &=\lim_{n\to\infty} (\sum_{i=1}^N a_i^n m(E_i))^{1/n} \end{align} $$

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].

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 fx(t)=f(x+t), gx(t)=g(t+x).

Then ||fx+f|| = ||fx+f+(gx+g)-(gx+g)||. By the triangle inequality, this is greater than or equal to ||gx+g||-||fx+f-(gx+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, |gx(t)+g(t)| is less than |gx(t)|+|g(t)| if and only if gx(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:


 * gx+g|| >= ||gx||+||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: limx->0||fx+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 gx and g have disjoint supports. Hence we can just say that the limit is ||gx||+||g||>=2||f||-2e.