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% Legendre transform
# 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, just
like 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 $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the
constant corresponding 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.
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