1. ## Legendre Transform

Hi guys

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

$
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
x<0: f*(p) = p2/4 + p -4

Now, for x=0 I get that f*(p) = 1 by the "usual" method.

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?

I would really appreciate any help.

2. 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

$
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.
..

3. Thanks for replying. Is there a rigorous way of showing that the interval for the Legendre Transform at x=0 has the domain p=[-2;2], or does it just have to be like this, since other values of p already have a transform associated with it?

4. Originally Posted by Niles_M
Thanks for replying. Is there a rigorous way of showing that the interval for the Legendre Transform at x=0 has the domain p=[-2;2], or does it just have to be like this, since other values of p already have a transform associated with it?
I did not give the domain to be [-2, 2]. I gave it to be p = 2, -2, that is, two values of p, not an interval. Look at the endpoints of the intervals for the other two transforms to get these values.

5. Originally Posted by mr fantastic
I did not give the domain to be [-2, 2]. I gave it to be p = 2, -2, that is, two values of p, not an interval. Look at the endpoints of the intervals for the other two transforms to get these values.
Hmm, I am a little unsure of why we are allowed to use x=(p+2)/2 and x=(p-2)/2 for x=0, when they are valid for x<0 and x>0, respectively. But in between (i.e. for -2<p<2), does this mean that the transformation is not defined for this interval?