Legendre transformation

File:Legendre transformation.png

The function <math>f(x)</math> is defined on the interval <math display="inline">[a,b]</math>. For a given <math>p</math>, the difference <math>px - f(x)</math> takes the maximum at <math>x'</math>. Thus, the Legendre transformation of <math>f(x)</math> is <math>f^*(p) =p x'-f(x')</math>.

In mathematics, the Legendre transformation (or Legendre transform), first introduced by Adrien-Marie Legendre in 1787 when studying the minimal surface problem,<ref name=":0">Template:Cite book</ref> is an involutive transformation on real-valued functions that are convex on a real variable. Specifically, if a real-valued multivariable function is convex on one of its independent real variables, then the Legendre transform with respect to this variable is applicable to the function.

In physical problems, the Legendre transform is used to convert functions of one quantity (such as position, pressure, or temperature) into functions of the conjugate quantity (momentum, volume, and entropy, respectively). In this way, it is commonly used in classical mechanics to derive the Hamiltonian formalism out of the Lagrangian formalism (or vice versa) and in thermodynamics to derive the thermodynamic potentials, as well as in the solution of differential equations of several variables.

For sufficiently smooth functions on the real line, the Legendre transform <math>f^*</math> of a function <math>f</math> can be specified, up to an additive constant, by the condition that the functions' first derivatives are inverse functions of each other. This can be expressed in Euler's derivative notation as <math display="block">Df(\cdot) = \left( D f^* \right)^{-1}(\cdot)~,</math> where <math>D</math> is an operator of differentiation, <math>\cdot</math> represents an argument or input to the associated function, <math>(\phi)^{-1}(\cdot)</math> is an inverse function such that <math>(\phi) ^{-1}(\phi(x))=x</math>, or equivalently, as <math>f'(f^{*\prime}(x^*)) = x^*</math> and <math>f^{*\prime}(f'(x)) = x</math> in Lagrange's notation.

The generalization of the Legendre transformation to affine spaces and non-convex functions is known as the convex conjugate (also called the Legendre–Fenchel transformation), which can be used to construct a function's convex hull.

Definition

Definition in one-dimensional real space

Let <math>I \sub \R</math> be an interval, and <math>f:I \to \R</math> a convex function; then the Legendre transform of <math>f</math> is the function <math>f^*:I^* \to \R</math> defined by <math display="block">f^*(x^*) = \sup_{x\in I}(x^*x-f(x)),\ \ \ \ I^*= \left \{x^*\in \R:\sup_{x\in I}(x^*x-f(x)) <\infty \right \}</math> where <math display="inline">\sup</math> denotes the supremum over <math>I</math>, e.g., <math display="inline">x</math> in <math display="inline">I</math> is chosen such that <math display="inline">x^*x - f(x)</math> is maximized at each <math display="inline">x^*</math>, or <math display="inline">x^*</math> is such that <math>x^*x-f(x)</math> has a bounded value throughout <math display="inline">I</math> (e.g., when <math>f(x)</math> is a linear function).

The function <math>f^*</math> is called the convex conjugate function of <math>f</math>. For historical reasons (rooted in analytic mechanics), the conjugate variable is often denoted <math>p</math>, instead of <math>x^*</math>. If the convex function <math>f</math> is defined on the whole line and is everywhere differentiable, then <math display="block">f^*(p)=\sup_{x\in I}(px - f(x)) = \left( p x - f(x) \right)|_{x = (f')^{-1}(p)} </math> can be interpreted as the negative of the <math>y</math>-intercept of the tangent line to the graph of <math>f</math> that has slope <math>p</math>.

Definition in n-dimensional real space

The generalization to convex functions <math>f:X \to \R</math> on a convex set <math>X \sub \R^n</math> is straightforward: <math>f^*:X^* \to \R</math> has domain <math display="block">X^*= \left \{x^* \in \R^n:\sup_{x\in X}(\langle x^*,x\rangle-f(x))<\infty \right \}</math> and is defined by <math display="block">f^*(x^*) = \sup_{x\in X}(\langle x^*,x\rangle-f(x)),\quad x^*\in X^* ~,</math> where <math>\langle x^*,x \rangle</math> denotes the dot product of <math>x^*</math> and <math>x</math>.

The Legendre transformation is an application of the duality relationship between points and lines. The functional relationship specified by <math>f</math> can be represented equally well as a set of <math>(x,y)</math> points, or as a set of tangent lines specified by their slope and intercept values.

Understanding the Legendre transform in terms of derivatives

For a differentiable convex function <math>f</math> on the real line with the first derivative <math>f'</math> and its inverse <math>(f')^{-1}</math>, the Legendre transform of <math>f</math>, <math> f^*</math>, can be specified, up to an additive constant, by the condition that the functions' first derivatives are inverse functions of each other, i.e., <math>f' = ((f^*)')^{-1}</math> and <math>(f^*)' = (f')^{-1}</math>.

To see this, first note that if <math> f</math> as a convex function on the real line is differentiable and <math> \overline{x} </math> is a critical point of the function of <math> x \mapsto p \cdot x -f(x) </math>, then the supremum is achieved at <math display="inline"> \overline{x}</math> (by convexity, see the first figure in this Wikipedia page). Therefore, the Legendre transform of <math> f</math> is <math> f^*(p)= p \cdot \overline{x} - f(\overline{x})</math>.

Then, suppose that the first derivative <math>f'</math> is invertible and let the inverse be <math> g = (f')^{-1} </math>. Then for each <math display="inline"> p</math>, the point <math> g(p)</math> is the unique critical point <math display="inline"> \overline{x}</math> of the function <math> x \mapsto px -f(x) </math> (i.e., <math> \overline{x} = g(p)</math>) because <math> f'(g(p))=p </math> and the function's first derivative with respect to <math>x</math> at <math> g(p)</math> is <math> p-f'(g(p))=0 </math>. Hence we have <math> f^*(p) = p \cdot g(p) - f(g(p))</math> for each <math display="inline"> p</math>. By differentiating with respect to <math display="inline"> p</math>, we find <math display="block">(f^*)'(p) = g(p)+ p \cdot g'(p) - f'(g(p)) \cdot g'(p).</math> Since <math> f'(g(p))=p</math> this simplifies to <math>(f^*)'(p) = g(p) = (f')^{-1}(p)</math>. In other words, <math>(f^*)'</math> and <math>f'</math> are inverses to each other.

In general, if <math> h' = (f')^{-1} </math> as the inverse of <math> f',</math> then <math> h' = (f^*)' </math> so integration gives <math> f^* = h +c.</math> with a constant <math> c. </math>

In practical terms, given <math>f(x),</math> the parametric plot of <math>xf'(x)-f(x)</math> versus <math>f'(x)</math> amounts to the graph of <math>f^*(p)</math> versus <math>p.</math>

In some cases (e.g. thermodynamic potentials, below), a non-standard requirement is used, amounting to an alternative definition of Template:Math with a minus sign, <math display="block">f(x) - f^*(p) = xp.</math>

Formal definition in physics context

In analytical mechanics and thermodynamics, Legendre transformation is usually defined as follows: suppose <math>f</math> is a function of <math>x</math>; then we have

<math>\mathrm{d} f = \frac{\mathrm{d} f}{\mathrm{d} x} \mathrm{d} x.</math>

Performing the Legendre transformation on this function means that we take <math>p = \frac{\mathrm{d} f}{\mathrm{d} x}</math> as the independent variable, so that the above expression can be written as

<math>\mathrm{d} f = p \mathrm{d} x,</math>

and according to Leibniz's rule <math>\mathrm{d} (uv) = u\mathrm{d} v + v\mathrm{d} u,</math> we then have

<math>\mathrm{d} \left(x p - f \right) = x \mathrm{d} p + p \mathrm{d} x - \mathrm{d} f = x\mathrm{d} p,</math>

and taking <math>f^* = xp-f,</math> we have <math>\mathrm d f^* = x \mathrm{d} p,</math> which means

<math>\frac{\mathrm{d} f^*}{\mathrm{d} p} = x.</math>

When <math>f</math> is a function of <math>n</math> variables <math>x_1, x_2, \cdots, x_n</math>, then we can perform the Legendre transformation on each one or several variables: we have

<math>\mathrm{d} f = p_1\mathrm{d} x_1 + p_2 \mathrm{d} x_2 + \cdots + p_n \mathrm{d} x_n,</math>

where <math>p_i = \frac{\partial f}{\partial x_i}.</math> Then if we want to perform the Legendre transformation on, e.g. <math>x_1</math>, then we take <math>p_1</math> together with <math>x_2, \cdots, x_n</math> as independent variables, and with Leibniz's rule we have

<math>\mathrm{d} (f - x_1 p_1) = -x_1 \mathrm{d} p_1 + p_2 \mathrm{d} x_2 + \cdots + p_n \mathrm{d} x_n.</math>

So for the function <math>\varphi(p_1, x_2, \cdots, x_n) = f(x_1, x_2, \cdots, x_n) - x_1 p_1,</math> we have

<math>\frac{\partial \varphi}{\partial p_1} = -x_1,\quad \frac{\partial \varphi}{\partial x_2} = p_2,\quad \cdots,

\quad \frac{\partial \varphi}{\partial x_n} = p_n.</math>

We can also do this transformation for variables <math>x_2, \cdots, x_n</math>. If we do it to all the variables, then we have

<math>\mathrm{d} \varphi = -x_1 \mathrm d p_1 - x_2 \mathrm{d} p_2 - \cdots - x_n \mathrm{d} p_n </math> where <math>\varphi = f-x_1 p_1 - x_2 p_2 - \cdots - x_n p_n. </math>

In analytical mechanics, people perform this transformation on variables <math>\dot q_1, \dot q_2, \cdots, \dot q_n </math> of the Lagrangian <math>L(q_1, \cdots, q_n, \dot{q}_1, \cdots, \dot{q}_n) </math> to get the Hamiltonian:

<math>H(q_1, \cdots, q_n, p_1, \cdots, p_n) = \sum_{i=1}^n p_i \dot{q}_i - L(q_1, \cdots, q_n, \dot{q}_1 \cdots, \dot{q}_n). </math>

In thermodynamics, people perform this transformation on variables according to the type of thermodynamic system they want; for example, starting from the cardinal function of state, the internal energy <math>U(S,V)</math>, we have

<math>\mathrm{d}U = T \mathrm{d} S - p \mathrm{d} V, </math>

so we can perform the Legendre transformation on either or both of <math>S, V </math> to yield

<math>\mathrm{d} H = \mathrm{d} (U + pV) \ \ \ \ \ \ \ \ \ \ = \ \ \ \ T\mathrm{d} S + V \mathrm{d} p</math>

<math>\mathrm{d} F = \mathrm{d} (U - TS) \ \ \ \ \ \ \ \ \ \ = -S\mathrm{d} T - p \mathrm{d} V</math>

<math>\mathrm{d} G = \mathrm{d} (U - TS + pV) = -S\mathrm{d} T + V \mathrm{d} p,</math>

and each of these three expressions has a physical meaning.

This definition of the Legendre transformation is the one originally introduced by Legendre in his work in 1787,<ref name=":0" /> and is still applied by physicists nowadays. Indeed, this definition can be mathematically rigorous if we treat all the variables and functions defined above: for example, <math>f,x_1,\cdots,x_n,p_1,\cdots,p_n, </math> as differentiable functions defined on an open set of <math>\R^n </math> or on a differentiable manifold, and <math>\mathrm{d} f, \mathrm{d} x_i, \mathrm{d} p_i </math> their differentials (which are treated as cotangent vector field in the context of differentiable manifold). This definition is equivalent to the modern mathematicians' definition as long as <math>f </math> is differentiable and convex for the variables <math>x_1, x_2, \cdots, x_n. </math>

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.Template:PbProof. 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.Template: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.)Template: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>).Template:Pb Note that <math>g</math> is also differentiable with the 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.Template: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, i.e., <math>f^{**} = f ~</math>.Template: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>.

Identities

As shown above, for a convex function <math>f(x)</math>, with <math>x = \bar{x}</math> maximizing or making <math>px - f(x)</math> bounded at each <math>p</math> to define the Legendre transform <math>f^*(p) = p\bar{x} - f(\bar{x})</math> and with <math>g \equiv (f')^{-1}</math>, the following identities hold.

• <math>f'(\bar{x}) = p</math>,
• <math>\bar{x} = g(p)</math>,
• <math>(f^*)'(p) = g(p)</math>.

Examples

Example 1

File:LegendreExample.svg

<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 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 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 domains 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 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 Template:Math defined on Template:Math, where Template:Math is a fixed constant.

For Template:Math fixed, the function of Template:Mvar, Template:Math has the first derivative Template:Math and second derivative Template:Math; there is one stationary point at Template:Math, which is always a maximum.

Thus, Template:Math and <math display="block">f^*(x^*)=\frac{ {x^*}^2}{4c} ~.</math>

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

Example 3

Let Template:Math for Template:Math.

For Template:Math fixed, Template:Math is continuous on Template:Mvar compact, hence it always takes a finite maximum on it; it follows that the domain of the Legendre transform of <math>f</math> is Template:Math.

The stationary point at Template:Math (found by setting that the first derivative of Template:Math with respect to <math>x</math> equal to zero) is in the domain Template:Math if and only if Template:Math. Otherwise the maximum is taken either at Template:Math or Template:Math because the second derivative of Template:Math 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 Template:Math 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 Template:Math is convex, for every Template:Mvar (strict convexity is not required for the Legendre transformation to be well defined). Clearly Template:Math is never bounded from above as a function of Template:Mvar, unless Template:Math. Hence Template:Math is defined on Template:Math and Template:Math. (The definition of the Legendre transform requires the existence of the supremum, that requires upper bounds.)

One may check involutivity: of course, Template:Math is always bounded as a function of Template:Math, hence Template:Math. Then, for all Template:Mvar one has <math display="block">\sup_{x^*\in\{c\}}(xx^*-f^*(x^*))=xc,</math> and hence Template:Math.

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 Template:Math, where Template:Mvar is a real, positive definite matrix.

Then Template:Mvar is convex, and <math display="block">\langle p,x\rangle-f(x)=\langle p,x \rangle-\langle x,Ax\rangle-c,</math> has gradient Template:Math and Hessian Template:Math, which is negative; hence the stationary point Template:Math is a maximum.

We have Template:Math, 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, Template:Math.

Let Template:Math be a function of two independent variables Template:Mvar and Template:Mvar, 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 Template:Mvar is convex in Template:Mvar for all Template:Mvar, so that one may perform the Legendre transform on Template:Mvar in Template:Mvar, with Template:Mvar the variable conjugate to Template:Mvar (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 Template:Mvar maximizing or making <math>px - f(x,y)</math> bounded for given Template:Mvar and Template:Mvar). Since the new independent variable of the transform with respect to Template:Mvar is Template:Mvar, the differentials Template:Math and Template:Math in Template:Mvar devolve to Template:Math and Template:Math in the differential of the transform, i.e., we build another function with its differential expressed in terms of the new basis Template:Math and Template:Math.

We thus consider the function Template:Math 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 Template:Math is the Legendre transform of Template:Math, where only the independent variable Template:Mvar has been supplanted by Template:Mvar. This is widely used in thermodynamics, as illustrated below.

Applications

Analytical mechanics

A Legendre transform is used in classical mechanics to derive the Hamiltonian formulation from the Lagrangian formulation, and conversely. A typical Lagrangian has the form

<math display="block">L(v,q)=\tfrac{1}2\langle v,Mv\rangle-V(q),</math> where <math>(v,q)</math> are coordinates on Template:Math, Template:Mvar is a positive definite real matrix, and <math display="block">\langle x,y\rangle = \sum_j x_j y_j.</math>

For every Template:Mvar fixed, <math>L(v, q)</math> is a convex function of <math>v</math>, while <math>V(q)</math> plays the role of a constant.

Hence the Legendre transform of <math>L(v, q)</math> as a function of <math>v</math> is the Hamiltonian function, <math display="block">H(p,q)=\tfrac {1}{2} \langle p,M^{-1}p\rangle+V(q).</math>

In a more general setting, <math>(v, q)</math> are local coordinates on the tangent bundle <math>T\mathcal M</math> of a manifold <math>\mathcal M</math>. For each Template:Mvar, <math>L(v, q)</math> is a convex function of the tangent space Template:Math. The Legendre transform gives the Hamiltonian <math>H(p, q)</math> as a function of the coordinates Template:Math of the cotangent bundle <math>T^*\mathcal M</math>; the inner product used to define the Legendre transform is inherited from the pertinent canonical symplectic structure. In this abstract setting, the Legendre transformation corresponds to the tautological one-form.Template:Explain

Thermodynamics

The strategy behind the use of Legendre transforms in thermodynamics is to shift from a function that depends on a variable to a new (conjugate) function that depends on a new variable, the conjugate of the original one. The new variable is the partial derivative of the original function with respect to the original variable. The new function is the difference between the original function and the product of the old and new variables. Typically, this transformation is useful because it shifts the dependence of, e.g., the energy from an extensive variable to its conjugate intensive variable, which can often be controlled more easily in a physical experiment.

For example, the internal energy Template:Mvar is an explicit function of the extensive variables entropy Template:Mvar, volume Template:Mvar, and chemical composition Template:Mvar (e.g., <math> i = 1, 2, 3, \ldots</math>) <math display="block"> U = U \left (S,V,\{N_i\} \right ),</math> which has a total differential <math display="block"> dU = T\,dS - P\,dV + \sum \mu_i \,dN _i</math>

where <math> T = \left. \frac{\partial U}{\partial S} \right \vert _{V, N_{i\ for\ all\ i\ values}}, P = \left. -\frac{\partial U}{\partial V} \right \vert _{S, N_{i\ for\ all\ i\ values}}, \mu_i = \left. \frac{\partial U}{\partial N_i} \right \vert _{S,V, N_{j\ for\ all\ j \ne i}}</math>.

(Subscripts are not necessary by the definition of partial derivatives but left here for clarifying variables.) Stipulating some common reference state, by using the (non-standard) Legendre transform of the internal energy Template:Mvar with respect to volume Template:Mvar, the enthalpy Template:Mvar may be obtained as the following.

To get the (standard) Legendre transform <math display="inline">U^*</math> of the internal energy Template:Mvar with respect to volume Template:Mvar, the function <math display="inline">u\left( p,S,V,\{{{N}_{i}}\} \right)=pV-U</math> is defined first, then it shall be maximized or bounded by Template:Mvar. To do this, the condition <math display="inline">\frac{\partial u}{\partial V} = p - \frac{\partial U}{\partial V} = 0 \to p = \frac{\partial U}{\partial V}</math> needs to be satisfied, so <math display="inline">U^* = \frac{\partial U}{\partial V}V - U</math> is obtained. This approach is justified because Template:Mvar is a linear function with respect to Template:Mvar (so a convex function on Template:Mvar) by the definition of extensive variables. The non-standard Legendre transform here is obtained by negating the standard version, so <math display="inline">-U^* = H = U - \frac{\partial U}{\partial V}V = U + PV</math>.

Template:Mvar is definitely a state function as it is obtained by adding Template:Mvar (Template:Mvar and Template:Mvar as state variables) to a state function <math display="inline"> U = U \left (S,V,\{N_i\} \right )</math>, so its differential is an exact differential. Because of <math display="inline"> dH = T\,dS + V\,dP + \sum \mu_i \,dN _i</math> and the fact that it must be an exact differential, <math> H = H(S,P,\{N_i\})</math>.

The enthalpy is suitable for description of processes in which the pressure is controlled from the surroundings.

It is likewise possible to shift the dependence of the energy from the extensive variable of entropy, Template:Mvar, to the (often more convenient) intensive variable Template:Mvar, resulting in the Helmholtz and Gibbs free energies. The Helmholtz free energy Template:Mvar, and Gibbs energy Template:Mvar, are obtained by performing Legendre transforms of the internal energy and enthalpy, respectively, <math display="block"> A = U - TS ~,</math><math display="block"> G = H - TS = U + PV - TS ~.</math>

The Helmholtz free energy is often the most useful thermodynamic potential when temperature and volume are controlled from the surroundings, while the Gibbs energy is often the most useful when temperature and pressure are controlled from the surroundings.

Variable capacitor

As another example from physics, consider a parallel conductive plate capacitor, in which the plates can move relative to one another. Such a capacitor would allow transfer of the electric energy which is stored in the capacitor into external mechanical work, done by the force acting on the plates. One may think of the electric charge as analogous to the "charge" of a gas in a cylinder, with the resulting mechanical force exerted on a piston.

Compute the force on the plates as a function of Template:Math, the distance which separates them. To find the force, compute the potential energy, and then apply the definition of force as the gradient of the potential energy function.

The electrostatic potential energy stored in a capacitor of the capacitance Template:Math and a positive electric charge Template:Math or negative charge Template:Math on each conductive plate is (with using the definition of the capacitance as <math display="inline">C = \frac{Q}{V}</math>),

<math display="block"> U (Q, \mathbf{x}) = \frac{1}{2} QV(Q,\mathbf{x}) = \frac{1}{2} \frac{Q^2}{C(\mathbf{x})},~</math>

where the dependence on the area of the plates, the dielectric constant of the insulation material between the plates, and the separation Template:Math are abstracted away as the capacitance Template:Math. (For a parallel plate capacitor, this is proportional to the area of the plates and inversely proportional to the separation.)

The force Template:Math between the plates due to the electric field created by the charge separation is then <math display="block"> \mathbf{F}(\mathbf{x}) = -\frac{dU}{d\mathbf{x}} ~. </math>

If the capacitor is not connected to any electric circuit, then the electric charges on the plates remain constant and the voltage varies when the plates move with respect to each other, and the force is the negative gradient of the electrostatic potential energy as <math display="block"> \mathbf{F}(\mathbf{x}) = \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}} \frac{Q^2}{{C(\mathbf{x})}^2} = \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}}V(\mathbf{x})^2 </math>

where <math display="inline"> V(Q,\mathbf{x}) = V(\mathbf{x}) </math> as the charge is fixed in this configuration.

However, instead, suppose that the voltage between the plates Template:Math is maintained constant as the plate moves by connection to a battery, which is a reservoir for electric charges at a constant potential difference. Then the amount of charges <math display="inline"> Q </math> is a variable instead of the voltage; <math display="inline"> Q </math> and <math display="inline"> V </math> are the Legendre conjugate to each other. To find the force, first compute the non-standard Legendre transform <math display="inline">U^*</math> with respect to <math display="inline"> Q </math> (also with using <math display="inline">C = \frac{Q}{V}</math>),

<math display="block">U^* = U - \left.\frac{\partial U}{\partial Q} \right|_\mathbf{x} \cdot Q =U - \frac{1}{2C(\mathbf{x})} \left. \frac{\partial Q^2}{\partial Q} \right|_\mathbf{x} \cdot Q = U - QV = \frac{1}{2} QV - QV = -\frac{1}{2} QV= - \frac{1}{2} V^2 C(\mathbf{x}).</math>

This transformation is possible because <math display="inline"> U </math> is now a linear function of <math display="inline"> Q </math> so is convex on it. The force now becomes the negative gradient of this Legendre transform, resulting in the same force obtained from the original function <math display="inline"> U </math>, <math display="block"> \mathbf{F}(\mathbf{x}) = -\frac{dU^*}{d\mathbf{x}} = \frac{1}{2} \frac{dC(\mathbf{x})}{d\mathbf{x}}V^2 .</math>

The two conjugate energies <math display="inline"> U </math> and <math display="inline"> U^* </math> happen to stand opposite to each other (their signs are opposite), only because of the linearity of the capacitance—except now Template:Math is no longer a constant. They reflect the two different pathways of storing energy into the capacitor, resulting in, for instance, the same "pull" between a capacitor's plates.

Probability theory

In large deviations theory, the rate function is defined as the Legendre transformation of the logarithm of the moment generating function of a random variable. An important application of the rate function is in the calculation of tail probabilities of sums of i.i.d. random variables, in particular in Cramér's theorem.

If <math>X_n</math> are i.i.d. random variables, let <math>S_n=X_1+\cdots+X_n</math> be the associated random walk and <math>M(\xi)</math> the moment generating function of <math>X_1</math>. For <math>\xi\in\mathbb R</math>, <math>E[e^{\xi S_n}] = M(\xi)^n</math>. Hence, by Markov's inequality, one has for <math>\xi\ge 0</math> and <math>a\in\mathbb R</math> <math display="block">P(S_n/n > a) \le e^{-n\xi a}M(\xi)^n=\exp[-n(\xi a - \Lambda(\xi))]</math> where <math>\Lambda(\xi)=\log M(\xi)</math>. Since the left-hand side is independent of <math>\xi</math>, we may take the infimum of the right-hand side, which leads one to consider the supremum of <math>\xi a - \Lambda(\xi)</math>, i.e., the Legendre transform of <math>\Lambda</math>, evaluated at <math>x=a</math>.

Microeconomics

Legendre transformation arises naturally in microeconomics in the process of finding the supply Template:Math of some product given a fixed price Template:Math on the market knowing the cost function Template:Math, i.e. the cost for the producer to make/mine/etc. Template:Math units of the given product.

A simple theory explains the shape of the supply curve based solely on the cost function. Let us suppose the market price for a one unit of our product is Template:Math. For a company selling this good, the best strategy is to adjust the production Template:Math so that its profit is maximized. We can maximize the profit <math display="block">\text{profit} = \text{revenue} - \text{costs} = PQ - C(Q)</math> by differentiating with respect to Template:Math and solving <math display="block">P - C'(Q_\text{opt}) = 0.</math>

Template:Math represents the optimal quantity Template:Math of goods that the producer is willing to supply, which is indeed the supply itself: <math display="block">S(P) = Q_\text{opt}(P) = (C')^{-1}(P).</math>

If we consider the maximal profit as a function of price, <math>\text{profit}_\text{max}(P)</math>, we see that it is the Legendre transform of the cost function <math>C(Q)</math>.

Geometric interpretation

For a strictly convex function, the Legendre transformation can be interpreted as a mapping between the graph of the function and the family of tangents of the graph. (For a function of one variable, the tangents are well-defined at all but at most countably many points, since a convex function is differentiable at all but at most countably many points.)

The equation of a line with slope <math>p</math> and <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. 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 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 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>

Legendre transformation in more than one dimension

For a differentiable real-valued function on an open convex subset Template:Mvar of Template:Math the Legendre conjugate of the pair Template:Math is defined to be the pair Template:Math, where Template:Mvar is the image of Template:Mvar under the gradient mapping Template:Math, and Template:Mvar is the function on Template:Mvar given by the formula <math display="block">g(y) = \left\langle y, x \right\rangle - f(x), \qquad x = \left(Df\right)^{-1}(y)</math> where <math display="block">\left\langle u,v\right\rangle = \sum_{k=1}^n u_k \cdot v_k</math>

is the scalar product on Template:Math. The multidimensional transform can be interpreted as an encoding of the convex hull of the function's epigraph in terms of its supporting hyperplanes.<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> This can be seen as consequence of the following two observations. On the one hand, the hyperplane tangent to the epigraph of <math>f</math> at some point <math>(\mathbf x, f(\mathbf x))\in U\times \mathbb{R}</math> has normal vector <math>(\nabla f(\mathbf x),-1)\in\mathbb{R}^{n+1}</math>. On the other hand, any closed convex set <math>C\in\mathbb{R}^m</math> can be characterized via the set of its supporting hyperplanes by the equations <math>\mathbf x\cdot\mathbf n = h_C(\mathbf n)</math>, where <math>h_C(\mathbf n)</math> is the support function of <math>C</math>. But the definition of Legendre transform via the maximization matches precisely that of the support function, that is, <math>f^*(\mathbf x)=h_{\operatorname{epi}(f)}(\mathbf x,-1) </math>. We thus conclude that the Legendre transform characterizes the epigraph in the sense that the tangent plane to the epigraph at any point <math>(\mathbf x,f(\mathbf x))</math> is given explicitly by<math display="block">\{\mathbf z\in\mathbb{R}^{n+1}: \,\, \mathbf z\cdot \mathbf x= f^*(\mathbf x)\}. </math>

Alternatively, if Template:Mvar is a vector space and Template:Math is its dual vector space, then for each point Template:Mvar of Template:Math and Template:Math of Template:Math, there is a natural identification of the cotangent spaces Template:Math with Template:Math and Template:Math with Template:Math. If Template:Mvar is a real differentiable function over Template:Math, then its exterior derivative, Template:Math, is a section of the cotangent bundle Template:Math and as such, we can construct a map from Template:Math to Template:Math. Similarly, if Template:Mvar is a real differentiable function over Template:Math, then Template:Math defines a map from Template:Math to Template:Math. If both maps happen to be inverses of each other, we say we have a Legendre transform. The notion of the tautological one-form is commonly used in this setting.

When the function is not differentiable, the Legendre transform can still be extended, and is known as the Legendre-Fenchel transformation. In this more general setting, a few properties are lost: for example, the Legendre transform is no longer its own inverse (unless there are extra assumptions, like convexity).

Legendre transformation on manifolds

Let <math display="inline">M</math> be a smooth manifold, let <math>E</math> and <math display="inline">\pi : E\to M</math> be a vector bundle on <math>M</math> and its associated bundle projection, respectively. Let <math display="inline">L : E\to \R</math> be a smooth function. We think of <math display="inline">L</math> as a Lagrangian by analogy with the classical case where <math display="inline">M = \R</math>, <math display="inline">E = TM = \Reals \times \Reals </math> and <math display="inline">L(x,v) = \frac 1 2 m v^2 - V(x)</math> for some positive number <math display="inline">m\in \Reals</math> and function <math display="inline">V : M \to \Reals</math>.

As usual, the dual of <math display="inline">E</math> is denote by <math display="inline">E^*</math>. The fiber of <math display="inline">\pi</math> over <math display="inline">x\in M</math> is denoted <math display="inline">E_x</math>, and the restriction of <math display="inline">L</math> to <math display="inline">E_x</math> is denoted by <math display="inline">L|_{E_x} : E_x\to \R</math>. The Legendre transformation of <math display="inline">L</math> is the smooth morphism<math display="block">\mathbf F L : E \to E^*</math> defined by <math display="inline">\mathbf FL(v) = d(L|_{E_x})_v \in E_x^*</math>, where <math display="inline">x = \pi(v)</math>. Here we use the fact that since <math display="inline">E_x</math> is a vector space, <math display="inline">T_v(E_x)</math> can be identified with <math display="inline">E_x</math>. In other words, <math display="inline">\mathbf FL(v)\in E_x^*</math> is the covector that sends <math display="inline">w\in E_x</math> to the directional derivative <math display="inline">\left.\frac d {dt}\right|_{t=0} L(v + tw)\in \R</math>.

To describe the Legendre transformation locally, let <math display="inline">U\subseteq M</math> be a coordinate chart over which <math display="inline">E</math> is trivial. Picking a trivialization of <math display="inline">E</math> over <math display="inline">U</math>, we obtain charts <math display="inline">E_U \cong U \times \R^r</math> and <math display="inline">E_U^* \cong U \times \R^r</math>. In terms of these charts, we have <math display="inline">\mathbf FL(x; v_1, \dotsc, v_r) = (x; p_1,\dotsc, p_r)</math>, where <math display="block">p_i = \frac {\partial L}{\partial v_i}(x; v_1, \dotsc, v_r)</math> for all <math display="inline">i = 1, \dots, r</math>. If, as in the classical case, the restriction of <math display="inline">L : E\to \mathbb R</math> to each fiber <math display="inline">E_x</math> is strictly convex and bounded below by a positive definite quadratic form minus a constant, then the Legendre transform <math display="inline">\mathbf FL : E\to E^*</math> is a diffeomorphism.<ref name="CdS2008">Ana Cannas da Silva. Lectures on Symplectic Geometry, Corrected 2nd printing. Springer-Verlag, 2008. pp. 147-148. Template:ISBN.</ref> Suppose that <math display="inline">\mathbf FL</math> is a diffeomorphism and let <math display="inline">H : E^* \to \R</math> be the "Hamiltonian" function defined by <math display="block">H(p) = p \cdot v - L(v),</math> where <math display="inline">v = (\mathbf FL)^{-1}(p)</math>. Using the natural isomorphism <math display="inline">E\cong E^{**}</math>, we may view the Legendre transformation of <math display="inline">H</math> as a map <math display="inline">\mathbf FH : E^* \to E</math>. Then we have<ref name="CdS2008"/> <math display="block">(\mathbf FL)^{-1} = \mathbf FH.</math>

Further properties

Scaling properties

The Legendre transformation has the following scaling properties: For Template:Math,

<math display="block">f(x) = a \cdot g(x) \Rightarrow f^\star(p) = a \cdot g^\star\left(\frac{p}{a}\right) </math> <math display="block">f(x) = g(a \cdot x) \Rightarrow f^\star(p) = g^\star\left(\frac{p}{a}\right).</math>

It follows that if a function is [[homogeneous function|homogeneous of degree Template:Mvar]] then its image under the Legendre transformation is a homogeneous function of degree Template:Mvar, where Template:Math. (Since Template:Math, with Template:Math, implies Template:Math.) Thus, the only monomial whose degree is invariant under Legendre transform is the quadratic.

Behavior under translation

<math display="block"> f(x) = g(x) + b \Rightarrow f^\star(p) = g^\star(p) - b</math> <math display="block"> f(x) = g(x + y) \Rightarrow f^\star(p) = g^\star(p) - p \cdot y </math>

Behavior under inversion

<math display="block"> f(x) = g^{-1}(x) \Rightarrow f^\star(p) = - p \cdot g^\star\left(\frac{1}{p} \right) </math>

Behavior under linear transformations

Let Template:Math be a linear transformation. For any convex function Template:Mvar on Template:Math, one has <math display="block"> (A f)^\star = f^\star A^\star </math> where Template:Math is the adjoint operator of Template:Mvar defined by <math display="block"> \left \langle Ax, y^\star \right \rangle = \left \langle x, A^\star y^\star \right \rangle, </math> and Template:Math is the push-forward of Template:Mvar along Template:Mvar <math display="block"> (A f)(y) = \inf\{ f(x) : x \in X , A x = y \}. </math>

A closed convex function Template:Mvar is symmetric with respect to a given set Template:Mvar of orthogonal linear transformations, <math display="block">f(A x) = f(x), \; \forall x, \; \forall A \in G </math> if and only if Template:Math is symmetric with respect to Template:Mvar.

Infimal convolution

The infimal convolution of two functions Template:Mvar and Template:Mvar is defined as

<math display="block"> \left(f \star_\inf g\right)(x) = \inf \left \{ f(x-y) + g(y) \, | \, y \in \mathbf{R}^n \right \}. </math>

Let Template:Math be proper convex functions on Template:Math. Then

<math display="block"> \left( f_1 \star_\inf \cdots \star_\inf f_m \right)^\star = f_1^\star + \cdots + f_m^\star. </math>

Fenchel's inequality

For any function Template:Mvar and its convex conjugate Template:Math Fenchel's inequality (also known as the Fenchel–Young inequality) holds for every Template:Math and Template:Math, i.e., independent Template:Math pairs, <math display="block">\left\langle p,x \right\rangle \le f(x) + f^\star(p).</math>

See also

• Dual curve
• Projective duality
• Young's inequality for products
• Convex conjugate
• Moreau's theorem
• Integration by parts
• Fenchel's duality theorem

References

• Template:Cite book
• Template:Cite book
• Fenchel, W. (1949). "On conjugate convex functions", Can. J. Math 1: 73-77.
• Template:Cite book
• Template:Cite journal

Further reading

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External links

• Legendre transform with figures at maze5.net
• Legendre and Legendre-Fenchel transforms in a step-by-step explanation at onmyphd.com