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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.)

ben
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2 Answers2

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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 $y = ax$.)

littleO
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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.

GReyes
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  • Thank you. To be clear, you say that the LT function with degree of homogeneity $q = \infty$ is considered convex. However, the same cannot be said of the original function with $r=1$, since, as you say at the start of your answer, convexity requires $r>1$ (and, I take it, concavity requires $0<r<1$). But then isn't this a problem? I remember reading somewhere that the LT can be used as a convexity test--i.e. if the LT of a function is concave then the original function must be convex. This seems to fail in the case of $r=1$, at least as you have described it above. – ben Jan 10 '20 at 17:53
  • Also, what do you mean by "your transform is zero at $p=$ slope of the line and infinity elsewhere."? By definition, $p=$ slope of the tangent line everywhere along the graph of $f$. There is no elsewhere, as far as $p$ is concerned. – ben Jan 10 '20 at 17:56
  • @ben The LT of a convex function is convex. For ex. $f(x)=x^2/2$ is its own transform and is convex. I may have been sloppy in my comment about the LT of a linear function $f(x)=ax$. Its transform is $f^(p)=0$ at $p=a$ (the slope of $f$) and is $f^(p)=+\infty$ at any other point. Such a function is by definition convex, all you need is convexity on its domain, which is just the point $p=a$. – GReyes Jan 11 '20 at 05:41
  • Ok. But the Legendre transform implies a one-to-one mapping between points and slopes, does it not? In the case of $f(x)=ax$, the slope at every point $x$ is constant ($a$). – ben Jan 16 '20 at 14:51