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Math Help - Legendre Transform

  1. #1
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    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

    <br />
f(x) = \left\{ {\begin{array}{*{20}c}<br />
 {x^2  + 1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x > 0}  \\<br />
   {x^2  + 1 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x < 0}  \\<br />
   {1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \,\,\,\,\,\,for\,\,\,\,x = 0}  \\<br />
\end{array}} \right.<br />

    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.
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  2. #2
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    Quote Originally Posted by Niles_M View Post
    Hi guys

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

    <br />
f(x) = \left\{ {\begin{array}{*{20}c}<br />
{x^2 + 1 + 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x > 0} \\<br />
{x^2 + 1 - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,\,\,x < 0} \\<br />
{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,for\,\,\,\,x = 0} \\<br />
\end{array}} \right.<br />

    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.
    ..
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by Niles_M View Post
    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.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    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?
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