The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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


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!

About these ads

August 26, 2010 - Posted by | Analysis, Measure Theory


  1. [...] We continue our project to show that the spaces are actually Banach spaces with Minkowski’s inequality. This will [...]

    Pingback by Minkowski’s Inequality « The Unapologetic Mathematician | August 27, 2010 | Reply

  2. [...] 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 , we’ve already [...]

    Pingback by The Supremum Metric « The Unapologetic Mathematician | August 30, 2010 | Reply

  3. [...] Banach Spaces To complete what we were saying about the spaces, we need to show that they’re complete. As it turns out, we can adapt the [...]

    Pingback by Some Banach Spaces « The Unapologetic Mathematician | August 31, 2010 | Reply

  4. [...] 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 . That is, not only do we know that for any [...]

    Pingback by The Extremal Case of Hölder’s Inequality « The Unapologetic Mathematician | September 1, 2010 | Reply

  5. [...] the continuous dual of for . That is, we’re excluding the case where either (but not its Hölder conjugate ) is infinite. And I say that when is -finite, the space of bounded linear functionals on is [...]

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

Leave a Reply

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

You are commenting using your 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


Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: