# 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 $x^p/p + y^q/q \ge xy\ \forall x,y > 0$. $\square$