% 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, analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)), 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.