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# 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

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

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 $$\{y\ns_1,\ldots,y\ns_n\}$$.