Jennifer Beasley

2022-02-15

Need help understanding the scaling properties of the Legendre transformation
At the wikipedia page for the Legendre transformation, there is a section on scaling properties where it says
$f\left(x\right)=ag\left(x\right)\to f\cdot \left(p\right)=ag\cdot \left(\frac{p}{a}\right)$
and $f\left(x\right)=g\left(ax\right)\to f\cdot \left(p\right)=g\cdot \left(\frac{p}{a}\right)$
where f*(p) and g*(p) are the Legendre transformations of f(x) and g(x), respectively, and a is a scale factor.
Also, it says:
"It follows that if a function is homogeneous of degree r then its image under the Legendre transformation is a homogeneous function of degree s, where $\frac{1}{r}+\frac{1}{s}=1$.
1) I don't see how the scaling properties hold. I'd appreciate if someone could spell this out for slow me.
2) I don't see how the relation between the degrees of homogeneity (r,s) follows from the scaling properties. Need some spelling out here too.
3) If the relation $\frac{1}{r}+\frac{1}{s}=1$ is true, then could this be used to prove that linearly homogeneous functions are not convex/concave? (Because convex/concave functions have a Legendre transformation, and $r=1$ would imply $s=\mathrm{\infty }$, which is absqrt and thus tantamount to saying that a function with $r=1$ has no Legendre transformation. No Legendre transformation then implies no convexity/concavity.)

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Heres

Call your two formulas (1) and (2). Suppose you have a homogeneous function f of degree $r>1$ (this is the range where it is convex). That means $f\left(kx\right)={k}^{r}f\left(x\right)$.
Now take the transform of both sides at p. Use (2) for the left hand side and (1) for the right hand side.
$f\cdot \left(\frac{p}{k}\right)={k}^{r}f\cdot \left({\frac{p}{k}}^{r}\right)$.
Now, if you call $\frac{p}{k}=s$ the above relation becomes
$f\cdot \left({k}^{1-r}s\right)={k}^{-r}f\cdot \left(s\right)$
Now, if you call , that is
$f\cdot \left(\lambda s\right)={\lambda }^{\frac{r}{r-1}}f\cdot \left(s\right)$
Threfore f* is homogeneous of degree $q=\frac{r}{r-1}$. It is straightforward to check that $\frac{1}{r}+\frac{1}{q}=1$. When $r=1$ your transform is zero at $p=$ slope of the line and is infinity elsewhere.You can interpret this formally as being homogeneous of infinite degree (both zero and infinity satisfy such homogeneity condition formally). A function like this is convex.

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