The Unapologetic Mathematician

Mathematics for the interested outsider

The Extremal Case of Hölder’s Inequality

We will soon need to know that Hölder’s inequality is in a sense the best we can do, at least for finite p. That is, not only do we know that for any f\in L^p and g\in L^q we have \lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q, but for any f\in L^p there is some g\in L^q for which we actually have equality. We will actually prove that

\displaystyle\lVert f\rVert_p=\max\left\{\left\lvert\int fg\,d\mu\right\rvert\Big\vert\lVert g\rVert_q\leq1\right\}

That is, not only is the integral bounded above by \lVert f\rVert_p\lVert g\rVert_q — and thus by \lVert f\rVert_p — but there actually exists some g in the unit ball which achieves this maximum.

Hölder’s inequality tells us that

\displaystyle\left\lvert\int fg\,d\mu\right\rvert\leq\int\lvert fg\rvert\,d\mu\leq\lVert f\rVert_p\lVert g\rVert_q\leq\lVert f\rVert_p

so \lVert f\rVert_p must be at least as big as every element of the given set. If \lVert f\rVert_p=0, then it’s clear that the asserted equality holds, since f=0 a.e., and so 0 is the only element of the set on the right. Thus from here we can assume \lVert f\rVert_p>0.

We now define a function g. At every point x where f(x)=0 we set g(x)=0 as well. At all other x we define

\displaystyle g(x)=\lVert f\rVert_p^{1-p}\frac{\lvert f(x)\rvert^p}{f(x)}

In the case where p=1 we will verify that \lVert g\rVert_\infty=1. That is, the essential supremum of g is 1. And, indeed, we find that g(x)=1 at points where f(x)>0, and g(x)=-1 at points where f(x)<0.

If 1<p<\infty, then we check

\displaystyle\begin{aligned}\lVert g\rVert_q&=\left(\int\lvert g\rvert^q\,d\mu\right)^\frac{1}{q}\\&=\left(\int\lVert f\rVert_p^{-p}\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\int\lvert f\rvert^p\,d\mu\right)^\frac{1}{q}\\&=\left(\lVert f\rVert_p^{-p}\lVert f\rVert_p^p\right)^\frac{1}{q}\\&=1\end{aligned}

In either case, it’s easy to see that

\displaystyle\int fg\,d\mu=\lVert f\rVert_p

as asserted.

About these ads

September 1, 2010 - Posted by | Analysis, Measure Theory

2 Comments »

  1. [...] is, is a bounded linear functional, and the operator norm is at most the norm of . The extremal case of Hölder’s inequality shows that there is some for which this is an equality, and thus we [...]

    Pingback by Some Continuous Duals « The Unapologetic Mathematician | September 3, 2010 | Reply

  2. I’m really thank you!
    You missed out one letter.
    “In the case where p=1, we will verify that ||g||inf = 1.”

    missed : 1

    Have a nice day!

    Comment by mathboy | April 20, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: