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 or the interesting article ‘Making Sense of the Legendre Transform’ available on the arxiv at

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