Editing Legendre transformation (section)

==Properties==
*The Legendre transform of a convex function, of which double derivative values are all positive, is also a convex function of which double derivative values are all positive.{{pb}}''Proof.'' Let us show this with a doubly differentiable function <math>f(x)</math> with all positive double derivative values and with a bijective (invertible) derivative.{{pb}} For a fixed <math>p</math>, let <math>\bar{x}</math> maximize or make the function <math>px - f(x)</math> bounded over <math>x</math>. Then the Legendre transformation of <math>f</math> is <math>f^*(p) = p\bar{x} - f(\bar{x})</math>, thus,<math display="block">f'(\bar{x}) = p</math>by the maximizing or bounding condition <math>\frac{d}{dx}(px - f(x)) = p - f'(x)= 0 </math>. Note that <math>\bar{x}</math> depends on <math>p </math>. (This can be visually shown in the 1st figure of this page above.){{pb}} Thus <math>\bar{x} = g(p)</math> where <math>g \equiv (f')^{-1}</math>, meaning that <math>g</math> is the inverse of <math>f'</math> that is the derivative of <math>f</math> (so <math>f'(g(p))= p</math>).{{pb}} Note that <math>g</math> is also differentiable with the [[Inverse functions and differentiation|following derivative (Inverse function rule)]],<math display="block">\frac{dg(p)}{dp} = \frac{1}{f''(g(p))} ~.</math>Thus, the Legendre transformation <math>f^*(p) = pg(p) - f(g(p))</math> is the composition of differentiable functions, hence it is differentiable.{{pb}} Applying the [[product rule]] and the [[chain rule]] with the found equality <math>\bar{x} = g(p)</math> yields<math display="block">\frac{d(f^{*})}{dp} = g(p) + \left(p - f'(g(p))\right)\cdot \frac{dg(p)}{dp} = g(p), </math>giving <math display="block">\frac{d^2(f^{*})}{dp^2} = \frac{dg(p)}{dp} = \frac{1}{f''(g(p))} > 0,</math>so <math>f^*</math> is convex with its double derivatives are all positive.
* The Legendre transformation is an [[Involution (mathematics)|involution]], i.e., <math>f^{**} = f ~</math>.{{pb}} ''Proof.'' By using the above identities as <math>f'(\bar{x}) = p</math>, <math>\bar{x} = g(p)</math>, <math>f^*(p) = p\bar{x} - f(\bar{x})</math> and its derivative <math>(f^*)'(p) = g(p)</math>, <math display="block">\begin{align}
f^{**}(y) &{} = \left(y\cdot \bar{p} - f^{*}(\bar{p})\right)|_{(f^{*})'(\bar{p}) = y} \\[5pt]
&{} = g(\bar{p})\cdot \bar{p} - f^{*}(\bar{p}) \\[5pt]
&{} = g(\bar{p})\cdot \bar{p} - (\bar{p} g(\bar{p})-f(g(\bar{p})))\\[5pt]
&{} = f(g(\bar{p})) \\[5pt]
&{} = f(y)~.
\end{align}</math>Note that this derivation does not require the condition to have all positive values in double derivative of the original function <math>f</math>.