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 $x$:

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 $L(f')$:

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

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