Need help understanding the scaling properties of the Legendre transformation

Question

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

$$ f(x)=ag(x) \rightarrow f^{\star}(p)=ag^{\star}(p/a)$$ and $$ f(x) = g(ax) \rightarrow f^{\star}(p) = g^{\star}(p/a)$$

where $f^{\star}(p)$ and $g^{\star}(p)$ 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 $1/r + 1/s = 1$."

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 $1/r + 1/s = 1$ 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 $s = \infty$, which is absurd and thus tantamount to saying that a function with $r=1$ has no Legendre transformation. No Legendre transformation then implies no convexity/concavity.)

Answer 1

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 $$ f(kx)=k^rf(x). $$ Now take the transform of both sides at $p$. Use (2) for the left hand side and (1) for the right hand side. $$ f^*(p/k)=k^rf^*(p/k^r). $$ Now, if you call $p/k=s$ the above relation becomes $$ f^*(k^{1-r}s)=k^{-r}f^*(s) $$ Now, if you call $k^{1-r}=\lambda$, then $k^{-r}=\lambda^{\frac{r}{r-1}}$, that is $$ f^*(\lambda s)=\lambda^{\frac{r}{r-1}}f^*(s) $$ Threfore $f^*$ is homogeneous of degree $q={\frac{r}{r-1}}$. It is straightforward to check that $1/r+1/q=1$. 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.

Answer 2

Here's an answer to your first question:

Suppose $a > 0$ and $f(x) = a g(x)$ for all $x \in \mathbb R^n$. Then \begin{align} f^*(z) &= \sup_x \, \langle z, x \rangle - f(x) \\ &= \sup_x \, \langle z, x \rangle - a g(x) \\ &= a \sup_x \, \langle z/a, x \rangle - g(x) \\ &= a g^*(z/a). \end{align}

Now suppose instead that $f(x) = g(ax)$ for all $x \in \mathbb R^n$. Then \begin{align} f^*(z) &= \sup_x \, \langle z, x \rangle - f(x) \\ &= \sup_x \, \langle z, x \rangle - g(ax) \\ \tag{1} &= \sup_y \, \langle z/a, y \rangle - g(y) \\ &= g^*(z/a). \end{align} (In step (1), we made a change of variable $y = ax$.)