Is L∞ Equivalent to Lp When p Approaches Infinity?

[EV33]

Homework Statement

My question is just on the definition of L^∞.

Is L^∞=L^p where p=∞, i.e.,

is a measurable function in L^∞ if ∫_Alf(x)l^∞<∞?

Homework Equations

*L^∞: The space of all bounded measurable functions on [0,1] (bounded except for possibly on a set of measure zero)

*A measurable function is said to belong to L^p if ∫_Alf(x)l^p<∞.

The Attempt at a Solution

Looks like it would be true based on the definition of L^p but I am really not sure since Royden only gives the one definition on L^∞.

 

[EV33]

Sorry, this should have have been posted in the calculus and beyond section.

 

[algebrat]

The infinity norm just take the maximum value of f. For general p norms, as p increases, it puts more weight on the larger terms. If you want an intuitive clue to the idea, notice that i you are measuring the hypotenuse based off of two values (the two sides), then if one of the sides is much larger than the other, than the hypotenuse comes pretty close to the length of the longer side. The infinity norm goes further, and simply gives you the length of the longer side, completely ignoring other lengths.

So does it make sense why L^infinity is all the bounded functions etc?

The problem with your expression with the integral of magnitude to the infinity, is when that is p, we really mean to take the pth root of the integral. So in the infinity norm, you would take the infinitieth root. So instead (they don't do that), they take the limit in p of the pth root, and prove that it is the supremum of |f|, but I might be forgetting a detail about sets of nonzero measure or something.

 

[Vargo]

So you need to take the limit of the p-norm as p goes to infty and show that this gives the "essential" supremum (i.e. the L infty norm). You prove this using the squeeze lemma.

Lets assume the measure of the entire space is finite for simplicity. I think the same arguments could be adapted, but suppose the measure of the whole space is M. According to Jensen's inequality applied to the convex function phi(x)=x^p, and using the original measure divided by M (giving a probability measure), we can prove that
\|f\|_p \leq M^{1/p}\|f\|_\infty
As long as M is not zero, then we find that the limsup of the p-norms is less than or equal to the L infinity norm.

For the other direction, let epsilon be positive. We know that there is a measurable subset E of positive measure on which the value of f is at least equal to \|f\|_\infty - \epsilon.

The Lp norm of f is bounded below by its Lp norm calculated over E.
\|f\|_p\geq \left( \int_E |f|^p d\mu\right)^{1/p}\geq \left( (\|f\|_\infty-\epsilon)^p\mu(E)\right)^{1/p}=(\|f\|_\infty-\epsilon)(\mu(E))^{1/p}

As long as the measure of E is positive, then we may conclude that the limsup of the Lp norm as p gets large is the L infinity norm.