legendre transform and convex function

[Mercury]

By "convex function", here, I mean a function whose first derivative
is strictly increasing.

Legendre transform applied to a convex function still gives a convex
function.

Please, can you give a proof (or an hint about it) of the statement ?

Warmest regards
mercury

[Ray Vickson]

On Nov 22, 8:20 am, Mercury <mercurynovam...@gmail.com> wrote:
> By "convex function", here, I mean a function whose first derivative
> is strictly increasing.

In Mathematics, convex means df/dx is non-decreasing. The concept you
are referring is called "strictly convex".

>
> Legendre transform applied to a convex function still gives a convex
> function.
>
> Please, can you give a proof (or an hint about it) of the statement ?
>
> Warmest regards
> mercury

The Legendre transform applied to f(x) produces a convex result f*(p),
no matter whether or not f(x) is convex. The proof is trivial: f*(p) =
max_{x} [p*x - f(x)]. For each p we are obtaining the max of several
linear (hence convex---but not strictly convex) functions, and it is
well-known that the max of a set of convex functions is itself convex.
(Note: in order to say this, we need to be careful about what we mean
by "convex"---hence, by insistence on differentiating between convex
and strictly convex in my opening remarks.)

Optimization people know about this under the name "duality" or the
"Fenchel dual" or the "Fenchel transform"; see
http://en.wikipedia.org/wiki/Convex_conjugate . In particular, look at
Theorem 2 of the link
http://www.maths.qmw.ac.uk/~ht/archive/lfth2.pdf , which is the second
external link in the cited Wiki page. The theorem is just stated
there, not proved; to prove it, look at
g = f*(a*p1 + (1-a)*p2), with 0 < a < 1 and p1 < p2. For fixed 'a' the
maximum over x that gives us g is at some x = x0; that is, g =
(a*p1+(1-a)*p2)*x0 - f(x0) = a*[p1*x0 - f(x0)] + (1-a)*[p2*x0 - f
(x0)]. Now p1*x0 - f(x0) <= f*(p1), because x0 might not be the
maximizer for x*p1-f(x), and similarly, p2*x0 - f(x) <= f*(p2). Thus,
we have f*(a*p1+(1-a)*p2) <= a*f*(p1) + (1-a)*f*(p2), so f* is a
convex function. NOTE: If f(x) is a strictly convex function then the
x giving the max is unique for each p; that means that p1*x0 - f(x0) <
f*(p1), etc., if 0 < a < 1 (strict inequality). That means that f*
(a*p1+(1-a)*p2) < a*f*(p1) + (1-a)*f*(p2) if 0 < a < 1, hence f* is
strictly convex. Examples show this need not happen if f is not
strictly convex.

R.G. Vickson