Editing Legendre transformation (section)

==Geometric interpretation==
For a [[strictly convex function]], the Legendre transformation can be interpreted as a mapping between the [[Graph of a function|graph]] of the function and the family of [[tangent]]s of the graph. (For a function of one variable, the tangents are well-defined at all but at most [[Countable set|countably many]] points, since a convex function is [[Derivative|differentiable]] at all but at most countably many points.)

The equation of a line with [[slope]] <math>p</math> and [[Y-intercept|<math>y</math>-intercept]] <math>b</math> is given by <math>y = p x + b</math>. For this line to be tangent to the graph of a function <math>f</math> at the point <math>\left(x_0, f(x_0)\right)</math> requires
<math display="block">f(x_0) = p x_0 + b</math>
and
<math display="block">p = f'(x_0).</math>

Being the derivative of a strictly convex function, the function <math>f'</math> is strictly monotone and thus [[Injective function|injective]]. The second equation can be solved for <math display="inline">x_0 = f^{\prime-1}(p),</math> allowing elimination of <math>x_0</math> from the first, and solving for the <math>y</math>-intercept <math>b</math> of the tangent as a function of its slope <math>p,</math> <math display="inline">b = f(x_0) - p x_0 = f\left(f^{\prime-1}(p)\right) - p \cdot f^{\prime-1}(p) = -f^\star(p)</math> where <math>f^{\star}</math> denotes the Legendre transform of <math>f.</math>

The [[Indexed family|family]] of tangent lines of the graph of <math>f</math> parameterized by the slope <math>p</math> is therefore given by
<math display="inline">y = p x - f^{\star}(p),</math> or, written implicitly, by the solutions of the equation
<math display="block">F(x,y,p) = y + f^{\star}(p) - p x = 0~.</math>

The graph of the original function can be reconstructed from this family of lines as the [[Envelope (mathematics)|envelope]] of this family by demanding
<math display="block">\frac{\partial F(x,y,p)}{\partial p} = f^{\star\prime}(p) - x = 0.</math>

Eliminating <math>p</math> from these two equations gives
<math display="block">y = x \cdot f^{\star\prime-1}(x) - f^{\star}\left(f^{\star\prime-1}(x)\right).</math>

Identifying <math>y</math> with <math>f(x)</math> and recognizing the right side of the preceding equation as the Legendre transform of <math>f^{\star},</math> yield <math display="inline">f(x) = f^{\star\star}(x) ~.</math>