User:Addemf/sandbox/Linfinity

The Relationship between Various Lp
I know that we have now accomplished all that we set out to do ... but ... one last thing.

It is natural to wonder if there is a relationship between $$L^p(E)$$ and $$L^q(E)$$ if $$1\le p< q$$. Indeed there is a simple relationship.

Moreover, it is natural to wonder if in some sense we might let p go to infinity, and if so, does anything good happen? Indeed we can, and indeed it does.

Let $$f(x) = 1/x$$. Show that $$f\in L^2([1,\infty))\smallsetminus L^1([1,\infty))$$.

Use this basic idea to show that for any $$1\le p < q$$ there is always a function in


 * $$L^q([1,\infty))\smallsetminus L^p([1,\infty))$$

In something like a converse, also show that there is always a function in


 * $$ L^p((0,1))\smallsetminus L^q((0,1))$$

The above exercise shows that, if there is to be some relationship between $$L^p(E)$$ and $$L^q(E)$$, it will not hold generally -- and it may depend on the nature of the subset E.

1. Let $$1\le p<q$$ and show that there exists some constant $$0<c$$ (which depends on p and q) such that for every $$f\in L^q((0,1))$$, $$\|f\|_p\le c\|f\|_q$$

Hint: Starting from the left side, write down a factor of 1 next to $$|f|^p$$ and apply the product bound. It may seem natural, when using the product bound, to take the p norm of $$|f|^p$$ but in fact this fails to produce any nice "cancellations".

Instead, prove that $$\||f|^p\|_q = \|f\|^p_q$$ and work through the remaining details.

2. Now argue that $$ L^q((0,1))\subset L^p((0,1))$$. Note that the containment is strict so you have two things to show: Both that the subset relation holds, and that there is some element in the set difference.

3. Finally, show how the above result can be generalized from the interval $$(0,1)$$ to some wider class of subsets of real numbers. At what point did you use any facts about $$(0,1)$$, and can you remove some details about this set which were unnecessary for the above proof?

L∞
As in the previous exercise, we will assume that $$\lambda(E)<\infty$$. (This is the generalization that you should have found at the end.

We now have some sense of what happens as we let p in $$L^p(E)$$ grow larger. In particular, you get an ever shrinking sequence of spaces of functions.

What then is the set


 * $$L^\infty(E) = \bigcap_{1\le p} L^p(E)$$

This is essentially the same question as finding a characterization of $$\lim_{p\to\infty}\|f\|_p$$.

1. Set $$E=(0,1)$$. Compute $$\lim_{p\to\infty}\|x\|_p$$.

Hint: If you're a little rusty with your tricky limits at infinity, use the "e to the ln" trick.

2. Set $$E=(0,a)$$ for some $$0<a$$. Compute $$\lim_{p\to\infty}\|x\|_p$$.

3. Set $$E=(0,2)$$. Compute $$\lim_{p\to\infty}\| 1-|x-1| \|_p$$.

4. What happens if you multiply any of these functions, in any of these examples, by a constant? What happens if you sum any of these with a constant?

Hint: Use the homogeneity of the norm for the constant multiple part -- don't make things harder than they have to be.

For the constant sum part, if you consider $$\sqrt[p]{\int_0^1(x+c)^p}$$ it will be helpful to consider the following:

Let $$0 \le b<a$$. Then


 * $$\lim_{p\to \infty}(a^p-b^p)^{1/p}=a$$

5. Based on all of the experimentation above, what do all of these values have in common?

The above exercise hopefully suggests that $$\|f\|_\infty$$ is the maximum of the function ... er, well, maybe not exactly the maximum of the function. After all, the very well-behaved function $$f(x)=x$$ on the interval (0,1) has no maximum.

Ok, so maybe $$\|f\|_\infty$$ is really the supremum rather than the maximum. Well, but even this doesn't work, because on a null set we can mess around with the function values and not change the value of the $$L^p(E)$$ norm.

Therefore we need a definition like this but which is not sensitive to values on any null set. Rather than talking about a maximum, or a supremum, we talk about an "essential supremum".

Let $$\lambda(E)<\infty$$ and let $$f:E\to\Bbb R$$ have a finite essential supremum, which we denote by $$\mathcal S$$.

1. Prove that $$\mathcal S\le\|f\|_\infty$$.

This will require using some analysis of integrals that we haven't done for a while. But it is natural to consider the indexed family of sets


 * $$A_t=\{ x\in E:|f(x)|\ge t\}$$ for any $$t\in [0, \mathcal S)$$

With this one may apply the ML bound to obtain the desired result.

2. Prove $$\|f\|_\infty\le \mathcal S$$.

For the reverse inequality, let $$\varepsilon\in\Bbb R^+$$ be arbitrarily small. Let


 * $$B_\varepsilon = \{x\in E:\mathcal S-\varepsilon < |f(x)| \}$$

Now prove that


 * $$\|f\|_p \le (\mathcal S-\varepsilon)\lambda(A_\varepsilon)^{1/p}$$

Then take the limits $$p\to\infty$$ and then $$\varepsilon\to 0$$.