---
title: "Legendre transform"
sort_title: "Legendre transform"
date: 2021-02-22
categories:
- Mathematics
- Physics
layout: "concept"
---

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 that $$L(f')$$ contains the same information as $$f(x)$$,
just in a different form,
analogously to e.g. the [Fourier transform](/know/concept/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 must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$:

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

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 fix 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:

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

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,
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](/know/concept/lagrangian-mechanics/)
and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other.
It is also used to convert between [thermodynamic potentials](/know/concept/thermodynamic-potential/).



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