## Hölder’s Inequality

We’ve seen the space of integrable functions on a measure space , which we called or . 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 , we define the space or to be the collection of all measurable functions for which is integrable. As in the case of , we identify two functions if they’re equal -almost everywhere.

It will turn out that these are Banach spaces. We define the norm

and we write to define a metric. This is clearly non-negative, and we see that if and only if -a.e., just as before. It’s also clear that . 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 is integrable if and are?

As a first step in this direction, we prove Hölder’s inequality: if and are real numbers greater than such that , and if and , then the product and . To see this, we will use the function defined for all positive real numbers by

Differentiating, we see that , so the only (positive) critical point of is . Since the limit as approaches and are both positive infinite, must be a local minimum. That is

For any two real numbers and , we can consider the value

and it follows that

and thus

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

Okay, so now we can turn to the theorem itself. If either or , the inequality clearly holds. Otherwise, we define

we can plug these into the above inequality to find

Since the measurability of and implies that of , and the right hand side of this inequality is integrable, we conclude that is integrable. If we integrate, we find

and Hölder’s inequality follows.

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

[…] 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 |

[…] 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 |

[…] 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 |

[…] 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 |

[…] 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 |

http://vankheakh.wordpress.com/2010/12/11/a-generalization-of-holders-inequality/

Comment by khea | December 18, 2010 |