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In order to guarantee the existence of the Legendre transform of a function $f:\mathbb{R}\rightarrow\mathbb{R}$, one usually has to know that $f$ is convex.

When performing a Legendre transformation on $U$ with respect to $S$ (obtaining the Helmholtz free energy) one is seemingly assuming that $U_{V,N}(S)$ is a convex function. Is this true? Why?

I know that $\frac{\partial U}{\partial S}=T$, so that $U$ is convex if $\frac{\partial T}{\partial S}>0$ (for all $S,V,N$). This certainly seems to be the case.

Qmechanic
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  • As a remark, upward directed convexity is a sufficient but not necessary condition. What you really need is that the derivative doesn't change its sign over the domain of interest. This guarantees that you can invert the functional dependencies to make a well-defined change of variable. – Phoenix87 Aug 14 '17 at 20:53
  • Possible duplicate of https://physics.stackexchange.com/q/176239/26129. See my answer there. – higgsss Aug 15 '17 at 13:22

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