2.15: Appendix II- Legendre Transformations

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A convex function of a single variable $f(x)$ is one for which $f''(x)>0$ everywhere. The Legendre transform of a convex function $f(x)$ is a function $g(p)$ defined as follows. Let $p$ be a real number, and consider the line $y=px$, as shown in Figure $\PageIndex{1}$. We define the point $x(p)$ as the value of $x$ for which the difference $F(x,p)=px-f(x)$ is greatest. Then define $g(p)=F\big(x(p),p\big)$.²⁶ The value $x(p)$ is unique if $f(x)$ is convex, since $x(p)$ is determined by the equation

$$f'\big(x(p)\big)=p\ .$$

Note that from $p=f'\big(x(p)\big)$ we have, according to the chain rule,

$${d\over dp}\,f'\big(x(p)\big)=f''\big(x(p)\big)\,x'(p)\qquad\Longrightarrow\qquad x'(p)=\Big[f''\big(x(p)\big)\Big]^{-1}\ .$$

From this, we can prove that $g(p)$ is itself convex:

$$\begin{split} g'(p)&={d\over dp} \Big[ p\,x(p)-f\big(x(p)\big)\Big]\\ &=p\,x'(p) + x(p) - f'\big(x(p)\big)\,x'(p)= x(p)\ , \end{split}$$

hence

$$g''(p)=x'(p)=\Big[f''\big(x(p)\big)\Big]^{-1}>0\ .$$

Figure $\PageIndex{1}$: Construction for the Legendre transformation of a function $f(x)$.

In higher dimensions, the generalization of the definition $f''(x)>0$ is that a function $F(x\ns_1,\ldots,x\ns_n)$ is convex if the matrix of second derivatives, called the Hessian,

$$H_{ij}(\Bx)={\pz^2 \!F\over\pz x\ns_i\,\pz x\ns_j}$$

is positive definite. That is, all the eigenvalues of $H_{ij}(\Bx)$ must be positive for every $\Bx$. We then define the Legendre transform $\BG(\Bp)$ as

$$\BG(\Bp)=\Bp\cdot\Bx-F(\Bx)$$

where

$$\Bp=\bnabla F\ .$$

Note that

$$dG=\Bx\cdot d\Bp + \Bp\cdot d\Bx - \bnabla F\cdot d\Bx = \Bx\cdot d\Bp\ ,$$

which establishes that $G$ is a function of $\Bp$ and that

$${\pz G\over\pz p\ns_j}=x\ns_j\ . \label{LTcond}$$

Note also that the Legendre transformation is self dual, which is to say that the Legendre transform of $G(\Bp)$ is $F(\Bx)$: $F\to G\to F$ under successive Legendre transformations.

We can also define a partial Legendre transformation as follows. Consider a function of $q$ variables $F(\Bx,\By)$, where $\Bx=\{x\ns_1,\ldots,x\ns_m\}$ and $\By=\{y\ns_1,\ldots,y\nd_n\}$, with $q=m+n$. Define $\Bp=\{p\ns_1,\ldots,p\ns_m\}$, and

$$G(\Bp,\By)=\Bp\cdot\Bx-F(\Bx,\By)\ ,$$

where

$$p\ns_a={\pz F\over \pz x\ns_a}\qquad (a=1,\ldots,m)\ .$$

These equations are then to be inverted to yield

$$x\ns_a=x\ns_a(\Bp,\By)={\pz G\over \pz p\ns_a}\ .$$

Note that

$$p\ns_a={\pz F\over \pz x\ns_a}\,\big(\Bx(\Bp,\By),\By\big)\ .$$

Thus, from the chain rule,

$$\delta\ns_{ab}={\pz p\ns_a\over\pz p\ns_b}={\pz^2\!F\over\pz x\ns_a\,\pz x\ns_c}\,{\pz x\ns_c\over\pz p\ns_b} ={\pz^2\!F\over\pz x\ns_a\,\pz x\ns_c}\,{\pz^2\!G\over\pz p\ns_c\,\pz p\ns_b}\ ,$$

which says

$${\pz^2 \! G\over\pz p\ns_a\,\pz p\ns_b}={\pz x\ns_a\over\pz p\ns_b}=\SK^{-1}_{ab}\ ,$$

where the $m\times m$ partial Hessian is

$${\pz^2\!F\over\pz x\ns_a\,\pz x\ns_b}={\pz p\ns_a\over\pz x\ns_b}=\SK\ns_{ab}\ .$$

Note that $\SK\ns_{ab}=\SK\ns_{ba}$ is symmetric. And with respect to the $\By$ coordinates,

$${\pz^2 \! G\over\pz y\ns_\mu\,\pz y\ns_\nu}=-{\pz^2 \! F\over\pz y\ns_\mu\,\pz y\ns_\nu}=-\SL\ns_{\mu\nu}\ ,$$

where

$$\SL\ns_{\mu\nu}={\pz^2\!F\over\pz y\ns_\mu\,\pz y\ns_\nu}$$

is the partial Hessian in the $\By$ coordinates. Now it is easy to see that if the full $q\times q$ Hessian matrix $H\ns_{ij}$ is positive definite, then any submatrix such as $\SK\ns_{ab}$ or $\SL\ns_{\mu\nu}$ must also be positive definite. In this case, the partial Legendre transform is convex in $\{p\ns_1,\ldots,p\ns_m\}$ and concave in $\{y\ns_1,\ldots,y\ns_n\}$.