Categories: Mathematics, Physics.

# Legendre transform

The Legendre transform of a function $$f(x)$$ is a new function $$L(f')$$, which depends only on the derivative $$f'(x)$$ of $$f(x)$$, and from which the original function $$f(x)$$ can be reconstructed. The point is, analogously to other transforms (e.g. Fourier), that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form.

Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of $$f(x)$$. Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$, which has a slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$-C$$:

\begin{aligned} y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C \end{aligned}

The Legendre transform $$L(f')$$ is defined such that $$L(f'(x_0)) = C$$ (or sometimes $$-C$$) for all $$x_0 \in [a, b]$$, where $$C$$ corresponds to the tangent line at $$x = x_0$$. This yields:

\begin{aligned} L(f'(x)) = f'(x) \: x - f(x) \end{aligned}

We want this function to depend only on the derivative $$f'$$, but currently $$x$$ still appears here as a variable. We fix that problem in the easiest possible way: by assuming that $$f'(x)$$ is invertible for all $$x \in [a, b]$$. If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is given by:

\begin{aligned} \boxed{ L(f') = f' \: x(f') - f(x(f')) } \end{aligned}

The only requirement for the existence of the Legendre transform is thus the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, which can only be true if $$f(x)$$ is either convex or concave, i.e. its derivative $$f'(x)$$ is monotonic.

Crucially, the derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$. In other words, the roles of $$f'$$ and $$x$$ are switched by the transformation: the coordinate becomes the derivative and vice versa. This is demonstrated here:

\begin{aligned} \boxed{ \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f') } \end{aligned}

Furthermore, Legendre transformation is an involution, meaning it is its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$:

\begin{aligned} g(L') = L' \: f'(L') - L(f'(L')) = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x) \end{aligned}

Moreover, the inverse of a (forward) transform always exists, because the Legendre transform of a convex function is itself convex. Convexity of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, which yields the following proof:

\begin{aligned} L''(f') = \dv{x(f')}{f'} = \dv{x}{f'(x)} = \frac{1}{f''(x)} > 0 \end{aligned}

Legendre transformation is important in physics, since it connects Lagrangian and Hamiltonian mechanics to each other. It is also used to convert between thermodynamic potentials.

1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.