Measure Theory/Linfinity

The Relationship between Various L^p

There is just one loose-end that we left dangling a couple lessons before, which is the issue of when $p=1$. There we asked whether it was possible for to make sense of the idea that the conjugate of p is $p^{*}=\mathrm{\infty }$, but for this to make sense we need to have some notion of $\Vert \cdot {\Vert }_{\mathrm{\infty }}$ and $L^{\mathrm{\infty }}(E)$.

It is also 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, if we let $p\to \mathrm{\infty }$ then do we discover a notion of $\Vert \cdot {\Vert }_{\mathrm{\infty }}$ which perhaps "plays nicely" with $\Vert \cdot {\Vert }_1$?

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

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

$L^q([1,\mathrm{\infty }))\setminus L^p([1,\mathrm{\infty }))$

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

$L^p((0,1))\setminus L^q((0,1))$

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 $f\in L^q((0,1))$,

  $\Vert f{\Vert }_p\le c\Vert f{\Vert }_q$

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 $\Vert |f{|}^p{\Vert }_q=\Vert f{\Vert }_q^p$ 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 $\lambda (E)<\mathrm{\infty }$. (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

$L^{\mathrm{\infty }}(E)=\bigcap _{1\le p}{L}^p(E)$

This is essentially the same question as finding a characterization of $\underset{p\to \mathrm{\infty }}{\lim }\Vert f{\Vert }_p$.

1. Set $E=(0,1)$.  Compute $\underset{p\to \mathrm{\infty }}{\lim }\Vert x{\Vert }_p$.  

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 $\underset{p\to \mathrm{\infty }}{\lim }\Vert x{\Vert }_p$.

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

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 $\sqrt[p]{\underset{0}{\overset{1}{\int }}(x+c{)}^p}$ it will be helpful to consider the following:

Let $0\le b<a$. Then

$\underset{p\to \mathrm{\infty }}{\lim }(a^p-b^p{)}^{1/p}=a$

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

The above exercise hopefully suggests that $\Vert f{\Vert }_{\mathrm{\infty }}$ 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 $\Vert f{\Vert }_{\mathrm{\infty }}$ 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 $\lambda (E)<\mathrm{\infty }$ and let $f:E\to ℝ$ have a finite essential supremum, which we denote by $𝒮$.

1. Prove that $𝒮\le \Vert f{\Vert }_{\mathrm{\infty }}$.

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 $t\in [0,𝒮)$

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

2. Prove $\Vert f{\Vert }_{\mathrm{\infty }}\le 𝒮$.

For the reverse inequality, let $\epsilon \in {ℝ}^{+}$ be arbitrarily small. Let

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

Now prove that

$\Vert f{\Vert }_p\le (𝒮-\epsilon )\lambda (A_{\epsilon }{)}^{1/p}$

Then take the limits $p\to \mathrm{\infty }$ and then $\epsilon \to 0$.

Show that the notion of an $L^p$ space can be extended to include $L^{\mathrm{\infty }}$. That is to say

1. Define the norm, space, and distance for $L^{\mathrm{\infty }}(E)$.

2. Prove that it is a vector space.

3. Prove that the concept of the conjugate of a real number $1\le p$ can be extended to include $p=\mathrm{\infty }$, and that the product-to-sum bound still applies.

4. Prove that the integral product bound does not hold but the triangle inequality still holds when $p=\mathrm{\infty }$, and that $L^{\mathrm{\infty }}(E)$ is complete. Be careful about any conditions which must hold in order for theorems to hold -- they may not be exactly the same conditions for finite p.