Editing Legendre transformation (section)

==Examples==

===Example 1===
[[Image:LegendreExample.svg|right|thumb|200px|<math> f(x) = e^x</math> over the domain <math>I=\mathbb{R}</math> is plotted in red and its Legendre transform <math>
f^*(x^*) = x^*(\ln(x^*) - 1)
</math> over the domain <math>I^* = (0, \infty)</math> in dashed blue. Note that the Legendre transform appears convex.]]
Consider the [[exponential function]] <math> f(x) = e^x,</math> which has the domain <math>I=\mathbb{R}</math>. From the definition, the Legendre transform is 
<math display="block">
f^*(x^*) = \sup_{x\in \mathbb{R}}(x^*x-e^x),\quad x^*\in I^*</math>
where <math>I^*</math> remains to be determined. To evaluate the [[Infimum and supremum|supremum]], compute the derivative of <math>x^*x-e^x</math> with respect to <math>x</math> and set equal to zero:
<math display="block">
\frac{d}{dx} (x^*x-e^x) = x^*-e^x = 0.
</math>
The [[Derivative_test#Second-derivative_test_(single_variable)|second derivative]] <math>-e^x</math> is negative everywhere, so the maximal value is achieved at <math>x = \ln(x^*)</math>. Thus, the Legendre transform is
<math display="block">
f^*(x^*) = x^*\ln(x^*)-e^{\ln(x^*)} = x^*(\ln(x^*) - 1)
</math>
and has domain <math>I^* = (0, \infty).</math> This illustrates that the [[domain of a function|domain]]s of a function and its Legendre transform can be different. 

To find the Legendre transformation of the Legendre transformation of <math> f</math>, 
<math display="block">
f^{**}(x) = \sup_{x^*\in \mathbb{R}}(xx^*-x^*(\ln(x^*) - 1)),\quad x\in I,
</math>
where a variable <math>
x
</math> is intentionally used as the argument of the function <math>
f^{**}
</math> to show the [[Involution (mathematics)|involution]] property of the Legendre transform as <math>
f^{**} = f
</math>. we compute
<math display="block">
\begin{aligned}
0 
&= \frac{d}{dx^*}\big( xx^*-x^*(\ln(x^*) - 1) \big)
= x - \ln(x^*)
\end{aligned}
</math>
thus the maximum occurs at <math>x^* = e^x</math> because the second derivative <math>
\frac{d^2}{{dx^*}^2}f^{**}(x) = - \frac{1}{x^*} < 0
</math> over the domain of <math>
f^{**}
</math> as <math>I^* = (0, \infty).</math> As a result, <math>
f^{**}
</math> is found as 
<math display="block">
\begin{aligned}
f^{**}(x)
&= xe^x - e^x(\ln(e^x) - 1) 
= e^x,
\end{aligned}
</math>
thereby confirming that <math>f = f^{**},</math> as expected.

===Example 2===
Let {{math|1=''f''(''x'') = ''cx''<sup>2</sup>}} defined on {{math|'''R'''}}, where {{math|''c'' > 0}} is a fixed constant.

For {{math|''x''*}} fixed, the function of {{mvar|x}}, {{math|1=''x''*''x'' − ''f''(''x'') = ''x''*''x'' − ''cx''<sup>2</sup>}} has the first derivative {{math|''x''* − 2''cx''}} and second derivative {{math|−2''c''}}; there is one stationary point at {{math|1=''x'' = ''x''*/2''c''}}, which is always a maximum.

Thus, {{math|1=''I''* = '''R'''}} and
<math display="block">f^*(x^*)=\frac{ {x^*}^2}{4c} ~.</math>

The first derivatives of {{math|''f''}}, 2{{math|''cx''}}, and of {{math|''f'' *}}, {{math|''x''*/(2''c'')}}, are inverse functions to each other. Clearly, furthermore,
<math display="block">f^{**}(x)=\frac{1}{4 (1/4c)}x^2=cx^2~,</math>
namely {{math|1=''f'' ** = ''f''}}.

===Example 3===
Let {{math|1=''f''(''x'') = ''x''<sup>2</sup>}} for {{math|1=''x'' ∈ (''I'' = [2, 3])}}.

For {{math|''x''*}} fixed, {{math|''x''*''x'' − ''f''(''x'')}} is continuous on {{mvar|I}} [[compact space|compact]], hence it always takes a finite maximum on it; it follows that the domain of the Legendre transform of <math>f</math> is {{math|1=''I''* = '''R'''}}.

The stationary point at {{math|1=''x'' = ''x''*/2}} (found by setting that the first derivative of {{math|''x''*''x'' − ''f''(''x'')}} with respect to <math>x</math> equal to zero) is in the domain {{math|[2, 3]}} if and only if {{math|4 ≤ ''x''* ≤ 6}}. Otherwise the maximum is taken either at {{math|1=''x'' = 2}} or {{math|1=''x'' = 3}} because the second derivative of {{math|''x''*''x'' − ''f''(''x'')}} with respect to <math>x</math> is negative as <math>-2</math>; for a part of the domain <math>x^* < 4</math> the maximum that {{math|''x''*''x'' − ''f''(''x'')}} can take with respect to <math>x \in [2,3]</math> is obtained at <math>x = 2</math> while for <math>x^* > 6</math> it becomes the maximum at <math>x = 3</math>. Thus, it follows that
<math display="block">f^*(x^*)=\begin{cases}
2x^*-4, & x^*<4\\
\frac{{x^*}^2}{4}, & 4\leq x^*\leq 6,\\
3x^*-9, & x^*>6.
\end{cases}</math>

===Example 4===
The function {{math|1=''f''(''x'') = ''cx''}} is convex, for every {{mvar|x}} (strict convexity is not required for the Legendre transformation to be well defined). Clearly {{math|1=''x''*''x'' − ''f''(''x'') = (''x''* − ''c'')''x''}} is never [[Bounded function|bounded from above]] as a function of {{mvar|x}}, unless {{math|1=''x''* − ''c'' = 0}}. Hence {{math|''f''*}} is defined on {{math|1=''I''* = {''c''}<nowiki/>}} and {{math|1=''f''*(''c'') = 0}}. ([[#Definition|The definition of the Legendre transform]] requires the existence of the [[Infimum and supremum|supremum]], that requires upper bounds.)

One may check involutivity: of course, {{math|''x''*''x'' − ''f''*(''x''*)}} is always bounded as a function of {{math|''x''*∈{''c''}<nowiki/>}}, hence {{math|1=''I''** = '''R'''}}. Then, for all {{mvar|x}} one has
<math display="block">\sup_{x^*\in\{c\}}(xx^*-f^*(x^*))=xc,</math>
and hence {{math|1=''f'' **(''x'') = ''cx'' = ''f''(''x'')}}.

=== Example 5 ===
As an example of a convex continuous function that is not everywhere differentiable, consider <math>f(x)= |x|</math>. This gives<math display="block">f^*(x^*) = \sup_{ x }(xx^*-|x|)=\max\left(\sup_{x\ge 0} x(x^*-1), 
 \,\sup_{x\le  0} x(x^*+1)  \right),</math>and thus <math>f^*(x^*)=0</math> on its domain <math>I^*=[-1,1]</math>.

===Example 6: several variables===
Let
<math display="block">f(x)=\langle x,Ax\rangle+c</math>
be defined on {{math|1=''X'' = '''R'''<sup>''n''</sup>}}, where {{mvar|A}} is a real, positive definite matrix.

Then {{mvar|f}} is convex, and
<math display="block">\langle p,x\rangle-f(x)=\langle p,x \rangle-\langle x,Ax\rangle-c,</math>
has gradient {{math|''p'' − 2''Ax''}} and [[Hessian matrix|Hessian]] {{math|−2''A''}}, which is negative; hence the stationary point {{math|1=''x'' = ''A''<sup>−1</sup>''p''/2}} is a maximum.

We have {{math|1=''X''* = '''R'''<sup>''n''</sup>}}, and
<math display="block">f^*(p)=\frac{1}{4}\langle p,A^{-1}p\rangle-c.</math>

==Behavior of differentials under Legendre transforms==
The Legendre transform is linked to [[integration by parts]], {{math|1=''p dx'' = ''d''(''px'') − ''x dp''}}.

Let {{math|''f''(''x'',''y'')}} be a function of two independent variables {{mvar|x}} and {{mvar|y}}, with the differential
<math display="block">df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy = p\,dx + v\,dy.</math>

Assume that the function {{mvar|f}} is convex in {{mvar|x}} for all {{mvar|y}}, so that one may perform the Legendre transform on {{mvar|f}} in {{mvar|x}}, with {{mvar|p}} the variable conjugate to {{mvar|x}} (for information, there is a relation <math>\frac{\partial f}{\partial x} |_{\bar{x}} = p</math> where <math>\bar{x}</math> is a point in {{mvar|x}} maximizing or making <math>px - f(x,y)</math> bounded for given {{mvar|p}} and {{mvar|y}}). Since the new independent variable of the transform with respect to {{mvar|f}} is {{mvar|p}}, the differentials {{math|''dx''}} and {{math|''dy''}} in {{mvar|df}} devolve to {{math|''dp''}} and {{math|''dy''}} in the differential of the transform, i.e., we build another function with its differential expressed in terms of the new basis {{math|''dp''}} and {{math|''dy''}}.

We thus consider the function {{math|1=''g''(''p'', ''y'') = ''f'' − ''px''}} so that
<math display="block">dg = df - p\,dx - x\,dp = -x\,dp + v\,dy</math>
<math display="block">x = -\frac{\partial g}{\partial p}</math>
<math display="block">v = \frac{\partial g}{\partial y}.</math>

The function {{math|−''g''(''p'', ''y'')}} is the Legendre transform of {{math|''f''(''x'', ''y'')}}, where only the independent variable {{mvar|x}} has been supplanted by {{mvar|p}}. This is widely used in [[thermodynamics]], as illustrated below.