I was reading the wiki article on Legendre Transform. I would be grateful if someone could explain the section at http://en.wikipedia.org/wiki/Legendre_transformation#Examples ie how they arrived at the Legendre transform of the quadratic form. Thanks.
Wikipedia Article -- Legendre Transform
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$\begingroup$
matrices
calculus-of-variations
vector-analysis
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0Would it help you if we replace "derivative" with "gradient" in the multivariate case? – 2011-08-14
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$f^*(y) = \sup_{x} \left(y^Tx - \frac1{2} x^TAx\right)$ The supremum can be found by taking the vector derivative with respect to $x$ and setting it equal to zero. $\frac{\partial \left(y^Tx \right)}{\partial x} = y$ and $\frac{\partial \left(x^T A x \right)}{\partial x} = \left( A+ A^T \right)x = 2Ax$ since $A$ is symmetric. Note that the supremum exists only if $A$ is positive definite. (second order condition) Setting the first derivative equal to zero gives us $x = A^{-1}y$ Plug this in to get $f^*(y) = \left(y^TA^{-1}y - \frac1{2} \left(A^{-1}y\right)^TAA^{-1}y\right) = \frac12 y^TA^{-1}y$
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0Thank you very much, Sivaram. – 2011-08-14