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 $f(x)$ can be reconstructed. The point is that $L(f')$ contains the same information as $f(x)$, just in a different form, analogously to e.g. the Fourier transform.

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 slope $f'(x_0)$ and intersects the $y$-axis at $y = -C$:

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

Where $C \equiv f'(x_0) \: x_0 - f(x_0)$. We now define the Legendre transform $L(f')$, such that for all $x_0 \in [a, b]$ we have $L(f'(x_0)) = C$ (some authors use $-C$ instead). Renaming $x_0$ to $x$:

\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 solve this 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\big(x(f')\big) } \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 is only satisfied if $f(x)$ is either convex or concave, meaning its derivative $f'(x)$ is monotonic.

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:

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

Furthermore, Legendre transformation is an involution, meaning it is its own inverse. To show this, 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, a Legendre transform is always invertible, because the transform of a convex function is itself convex. Convexity of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, so a proof is:

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

And an analogous proof exists for concave functions where $f''(x) < 0$.

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.

## References

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