# The Unapologetic Mathematician

## Hölder’s Inequality

We’ve seen the space of integrable functions on a measure space $X$, which we called $L^1$ or $L^1(\mu)$. We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to generalize.

Given a real number $p>1$, we define the space $L^p$ or $L^p(\mu)$ to be the collection of all measurable functions $f$ for which $\lvert f\rvert^p$ is integrable. As in the case of $L^1$, we identify two functions if they’re equal $\mu$-almost everywhere.

It will turn out that these are Banach spaces. We define the $L^p$ norm $\displaystyle\lVert f\rVert_p=\left(\int\lvert f\rvert^p\,d\mu\right)^{\frac{1}{p}}$

and we write $\rho_p(f,g)=\lVert f-g\rVert_p$ to define a metric. This is clearly non-negative, and we see that $\rho_p(f,g)=0$ if and only if $f=g$ $\mu$-a.e., just as before. It’s also clear that $\lVert cf\rVert_p=\lvert c\rvert\lVert f\rVert_p$. What we need to work to check is the triangle inequality. It’s also not quite so apparent a problem, but we actually don’t know yet that this is a vector space at all! That is, how do we know that $\lvert f-g\rvert^p$ is integrable if $\lvert f\rvert^p$ and $\lvert g\rvert^p$ are?

As a first step in this direction, we prove Hölder’s inequality: if $p$ and $q$ are real numbers greater than $1$ such that $\frac{1}{p}+\frac{1}{q}=1$, and if $f\in L^p$ and $g\in L^q$, then the product $fg\in L^1$ and $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$. To see this, we will use the function $\phi$ defined for all positive real numbers by $\displaystyle\phi(t)=\frac{t^p}{p}+\frac{t^{-q}}{q}$

Differentiating, we see that $\phi'(t)=t^{p-1}-t^{-q-1}=t^{-q-1}(t^{p+q}-1)$, so the only (positive) critical point of $\phi$ is $1$. Since the limit as $t$ approaches $0$ and $\infty$ are both positive infinite, $t=1$ must be a local minimum. That is $\displaystyle\frac{t^p}{p}+\frac{t^{-q}}{q}=\phi(t)\geq\phi(1)=\frac{1}{p}+\frac{1}{q}=1$

For any two real numbers $a$ and $b$, we can consider the value $\displaystyle t=\frac{a^\frac{1}{q}}{b^\frac{1}{p}}$

and it follows that \displaystyle\begin{aligned}1&\leq\phi(t)\\&=\frac{t^p}{p}+\frac{t^{-q}}{q}\\&=\frac{\frac{a^\frac{p}{q}}{b}}{p}+\frac{\frac{a^{-1}}{b^{-\frac{q}{p}}}}{q}\\&=\frac{a^\frac{p}{q}}{bp}+\frac{b^\frac{q}{p}}{aq}\\&=\frac{a^{p-1}}{bp}+\frac{b^{q-1}}{aq}\end{aligned}

and thus $\displaystyle ab\leq\frac{a^p}{p}+\frac{b^q}{q}$

which is clearly also true even if we allow $a$ or $b$ to be zero. This is known as “Young’s inequality”.

Okay, so now we can turn to the theorem itself. If either $\lVert f\rVert_p=0$ or $\lVert g\rVert_q=0$, the inequality clearly holds. Otherwise, we define $\displaystyle a=\frac{\lvert f\rvert}{\lVert f\rVert_p}\qquad b=\frac{\lvert g\rvert}{\lVert g\rVert_q}$

we can plug these into the above inequality to find \displaystyle\begin{aligned}\frac{\lvert fg\rvert}{\lVert f\rVert_p\lVert g\rVert_q}&=\frac{\lvert f\rvert}{\lVert f\rVert_p}\frac{\lvert g\rvert}{\lVert g\rVert_q}\\&\leq\frac{\lvert f\rvert^p}{p\lVert f\rVert_p^p}+\frac{\lvert g\rvert^q}{q\lVert g\rVert_q^q}\\&=\frac{1}{p}\frac{\lvert f\rvert^p}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\lvert g\rvert^q}{\int\lvert g\rvert^q\,d\mu}\end{aligned}

Since the measurability of $f$ and $g$ implies that of $\lvert fg\rvert$, and the right hand side of this inequality is integrable, we conclude that $\lvert fg\rvert$ is integrable. If we integrate, we find $\displaystyle\frac{\lVert fg\rVert_1}{\lVert f\rVert_p\lVert g\rVert_q}\leq\frac{1}{p}\frac{\int\lvert f\rvert^p\,d\mu}{\int\lvert f\rvert^p\,d\mu}+\frac{1}{q}\frac{\int\lvert g\rvert^q\,d\mu}{\int\lvert g\rvert^q\,d\mu}=\frac{1}{p}+\frac{1}{q}=1$

and Hölder’s inequality follows.

The condition relating $p$ and $q$ is very common in this discussion, so we will say that such a pair of real numbers are “Hölder conjugates” of each other. Given $p$, the Hölder conjugate $q$ is uniquely defined by $q=\frac{p}{p-1}$, which is a strictly decreasing function sending $(1,\infty)$ to itself (with order reversed, of course). The fact that this function has a (unique) fixed point at $2$ will be important. In particular, we will see that this norm is associated with an inner product on $L^2$, and that Hölder’s inequality actually implies the Cauchy-Schwarz inequality!

August 26, 2010 Posted by | Analysis, Measure Theory | 6 Comments