Originally Posted by

**Niles_M** Hi guys

I am looking at f(x) = (|x|+1)2. I have to find its Legendre transform, and thus I write this as

$\displaystyle

f(x) = \left\{ {\begin{array}{*{20}c}

{x^2 + 1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x > 0} \\

{x^2 + 1 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x < 0} \\

{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,for\,\,\,\,x = 0} \\

\end{array}} \right.

$

I use the fact that for x>0 and x<0 the function is strictly convex, so for these two intervals the Legendre Transform is

x>0: f*(p) = p2/4 - p Mr F says: for p > 2 (since x = (p - 2)/2 for x > 0 therefore x > 0 => p > 2).

x<0: f*(p) = p2/4 + p -4 Mr F says: I get f*(p) = p^2/4 + p for p < -2 (since x = (p + 2)/2 therefore x < 0 => p < -2).

Now, for x=0 I get that f*(p) = 1 by the "usual" method. Mr F says: I get f*(p) = -1 for p = -2, 2.

My question is now: What is the next step? Does this mean that the function f(x) has 3 Legendre Transforms that all work for all p? Mr F says: Each transform has a different domain.

I would really appreciate any help.