The Legendre Transform

This topic doesn’t get anywhere near the attention it deserves. I like it because it leads to a useful family of inequalities. Given a strictly convex function f\colon\mathbb{R}\to\mathbb{R} (i.e. f'' > 0 and f' is monotone) we define its Legendre transform as

\displaystyle g(y) = \max_{x\in\mathbb{R}}\left\{xy - f(x)\right\}.

The maximum can be calculated by differentiating and requiring that \frac{d}{dx}(xy-f(x))=0. Hence we can write y = y(x) := f'(x) and use the fact that this function is invertible to write x = x(y). With these we can check that this was indeed a maximisation because \frac{d^2}{dx^2}(xy-f(x))=-f''(x) < 0 by assumption.

Now, with y = y(x) = y(x(y)) as above we get the Legendre transform:

g(y) = y x(y) - f(x(y)).

Omitting the display of the functional dependency we get the symmetric relationship g(y) + f(x) = xy and this suggests (correctly) that the Legendre transform is its own inverse.

For more information you can check out the wikipedia page at http://en.wikipedia.org/wiki/Legendre_transform or the interesting article ‘Making Sense of the Legendre Transform’ available on the arxiv at http://arxiv.org/abs/0806.1147.

To finish we’ll use the Legendre transform to prove the following useful inequality.

Proposition. If p,q\in(1,\infty) satisfy \frac{1}{p} + \frac{1}{q} = 1 then xy \le \frac{x^p}{p} + \frac{y^q}{q} for all real x,y > 0.

Proof. Define f(x) = x^p/p and then y(x) = x^{p-1} gives x(y) = y^{q-1} since it is readily shown that (p-1)(q-1)=1. Therefore g(y) = y x(y) - f(x(y)) = y^q - (y^{q-1})^p/p = y^q(1-1/p) = y^q/q since (q-1)p = q. Therefore, by definition, g(y) = \max_x\{xy - f(x)\}\ge xy-f(x) implies that x^p/p + y^q/q \ge xy\ \forall x,y > 0. \square