The Unapologetic Mathematician

Mathematics for the interested outsider

The Supremum Metric

We can actually extend what we’ve been doing with Hölder’s inequality and Minkowski’s inequality a little further. Given a metric space (X,\mathcal{S},\mu), we’ve already discussed the idea of an “essentially bounded” function — one for which there is some real constant c so that f(x)\leq c for almost all x\in X. We will write L^\infty(X) for the collection of essentially bounded functions on the measure space. It should be clear that these form a vector space.

We also discussed the “essential supremum” \text{ess sup}(\lvert f\rvert) of an essentially bounded function. We’ll now write this as \lVert f\rVert_\infty, suggesting that it’s a norm. And it’s clear that \lVert cf\rVert_\infty=\lvert c\rvert\lVert f\rVert_\infty, and that \lVert f\rVert_\infty=0 if and only if f=0 almost everywhere. Verifying the triangle identity is exactly Minkowski’s inequality.

And, indeed, we know that \lvert f(x)\rvert\leq\lVert f\rVert_\infty and \lvert g(x)\rvert\leq\lVert g\rVert_\infty a.e., so \lvert f(x)+g(x)\rvert\leq\lvert f(x)\rvert+\lvert g(x)\rvert\leq\lVert f\rVert_\infty+\lVert g\rVert_\infty a.e., so whatever the least such essential upper bound is smaller still. That is, \lVert f+g\rVert_\infty=\lVert f\rVert_\infty+\lVert g\rVert_\infty.

Now for Hölder’s inequality. For this purpose we consider \frac{1}{\infty}=0, and thus \frac{1}{1}+\frac{1}{\infty}=1, which means that 1 and \infty are Hölder-conjugates. Thus our assertion is that if f is integrable and g is essentially bounded, then fg is integrable and \lVert fg\rVert_1\leq\lVert f\rVert_1\lVert g\rVert_\infty. Indeed, we know that \lvert g(x)\rvert\leq\lVert g\rVert_\infty, and so \lvert f(x)g(x)\rvert=\lvert f(x)\rvert\lvert g(x)\rvert\leq \lvert f(x)\rvert\lVert g\rVert_\infty — both inequalities holding almost everywhere. From this, we conclude that

\displaystyle\begin{aligned}\lVert fg\rVert_1&=\int\lvert fg\rvert\,d\mu\\&\leq\int\lvert f\rvert\lVert g\rVert_\infty\,d\mu\\&=\int\lvert f\rvert\,d\mu\lVert g\rVert_\infty\\&=\lVert f\rVert_1\lVert g\rVert_\infty\end{aligned}

as we asserted. From now on, we’ll allow p=\infty (and q=1) whenever we’re talking about a Hölder-conjugate pair or L^p-space.

August 30, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. […] we’ll show that the sequence converges in the supremum norm on . That is, we’ll show that there is some so that the maximum of the difference for […]

    Pingback by The Picard Iteration Converges « The Unapologetic Mathematician | May 6, 2011 | Reply

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