Legendre transform (1 Viewer)

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Hi guys

I am looking at f(x) = (|x|+1)2. I write this as

[tex]
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.
[/tex]

I want to find the Legendre transform of this function. For x>0 I get the Legendre transform

f*(p) = p2/4-p-1/2.

I am a little unsure of how this works. Because I need to find the Legendre transform of f for x<0 and x=0. But how do these solutions get "patched" together?
 

HallsofIvy

Science Advisor
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I don't get that at all. For x> 0, f is differentiable so f*(p) is just -f(x) where x is such that f'(x)= p. Here, f'(x)= 2x+ 2= p when x= (p-2)/2. So f*(p)= -f((p-2)/2)= -((p-2)2/4+ 1+ 2(p-1)/2)= -((1/4)p2- p+ 1+ 1+ p- 2)= -(1/4)p^2.

Similarly, for x< 0, f'(x)= 2x- 2= p when x= (p+2)/2. So F*(p)= -f((p+2)/2)= -)(p+2)2/4+ 1+ 2(p+2)/2)= -((1/4)p2+ p+ 1+ 1+ p+ 2)= -(1/4)p2- 2p- 4.

f is not differentiable at x= 0 so have to go back to the basic definition: f*(0)= max(0(x)-f(x))= max -f(x). Of course, that maximum is 0: f*(0)= 0. Since f is not differentiable at 0, it Legendre transform is not continuous there and they cannot be "patched" together.
 
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Thanks. Ok, I've done it a slight different way. First, I must admit that I have not heard of a theorem stating that if f is differentiable, then f*(p) is just -f(x).

I used that since f is strictly convex (we look at x>0 for now), then f*(p)=xp-f(x(p)), where x is such that f'(x)= p.


But OK: Let us look at the solutions found in #2 (I am mostly trying to learn the method). So

for x>0: f*(p) = -(1/4)p^2
for x<0: f*(p) = -(1/4)p^2- 2p- 4
x=0 : f*(p) = 0 (by the way, shouldn't this be -1 rather than 0?)


What solutions are "valid" for the different intervals of p? I mean, we have 3 solutions, one for each interval of x. But how do I find out what intervals of p they correspond to?
 
Last edited:

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