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.)

At the wikipedia page for the Legendre transformation, there is a section on scaling properties where it says

and

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

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

utripljigmp

Beginner2022-02-16Added 12 answers

Heres

maskorfclp

Beginner2022-02-17Added 12 answers

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$k}^{1-r}=\lambda ,\text{then}\text{}{k}^{-r}={\lambda}^{\frac{r}{r-1}$ , 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.

Now take the transform of both sides at p. Use (2) for the left hand side and (1) for the right hand side.

Now, if you call

Now, if you call

Threfore f* is homogeneous of degree

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