Legendre Transformation

Legendre transform for a given function $f(x)$ ${\ displaystyle f (x)}$ $f (x)$ Is the construction of a function $f^{*}(p)$ ${\ displaystyle f ^ {*} (p)}$ ${\ displaystyle f ^ {*} (p)}$ dual to her according to Jung. If the original function was defined on a vector space $V$ ${\ displaystyle V}$ $V$ , its Legendre transformation will be a function defined on the adjoint space $V^{*}$ ${\ displaystyle V ^ {*}}$ $V ^ *$ , i.e., on the space of linear functionals on the space $V$ ${\ displaystyle V}$ $V$ .

Motivation

Possible motivation can be expressed as a less general definition. The Legendre transformation is such a replacement of a function and a variable in which the old derivative is taken as a new variable, and the old variable is taken as a new derivative.

Differential expression

df(x)=f^{\prime }(x)dx

{\ displaystyle df (x) = f ^ {\ prime} (x) dx}

df(x)=f^{\prime }(x)dx

due to the fact that $d(xf^{\prime })=f^{\prime }dx+xdf^{\prime }$ ${\ displaystyle d (xf ^ {\ prime}) = f ^ {\ prime} dx + xdf ^ {\ prime}}$ $d(xf^{\prime })=f^{\prime }dx+xdf^{\prime }$ can be written as

d(xf^{\prime }-f)=xdf^{\prime }.

{\ displaystyle d (xf ^ {\ prime} -f) = xdf ^ {\ prime}.}

d(xf^{\prime }-f)=xdf^{\prime }.

If we now accept that

F=xf^{\prime }-f~,~~y=f^{\prime }(x),

{\ displaystyle F = xf ^ {\ prime} -f ~, ~~ y = f ^ {\ prime} (x),}

F=xf^{\prime }-f~,~~y=f^{\prime }(x),

which is the transformation of Legendre $(f,x)\to (F,y)$ ${\ displaystyle (f, x) \ to (F, y)}$ $(f,x)\to (F,y)$ then

dF(y)=F^{\prime }(y)dy~,~x=F^{\prime }(y).

{\ displaystyle dF (y) = F ^ {\ prime} (y) dy ~, ~ x = F ^ {\ prime} (y).}

dF(y)=F^{\prime }(y)dy~,~x=F^{\prime }(y).

With this new variable $y$ ${\ displaystyle y}$ $y$ equal to the old derivative, and the old variable $x$ ${\ displaystyle x}$ $x$ equal to the new derivative:

y=f^{\prime }(x)~,~x=F^{\prime }(y).

{\ displaystyle y = f ^ {\ prime} (x) ~, ~ x = F ^ {\ prime} (y).}

y=f^{\prime }(x)~,~x=F^{\prime }(y).

Definitions may differ in sign. $F$ ${\ displaystyle F}$ $F$ . If the source variables $x$ ${\ displaystyle x}$ $x$ more than one, the Legendre transformation can be carried out on any subset of them.

Definition

Analytical Definition

Legendre conversion function $f$ ${\ displaystyle f}$ $f$ defined on a subset $M$ ${\ displaystyle M}$ $M$ vector space $V$ ${\ displaystyle V}$ $V$ is called a function $f^{*}$ ${\ displaystyle f ^ {*}}$ $f^{*}$ defined on a subset $M^{*}$ ${\ displaystyle M ^ {*}}$ $M^*$ conjugate space $V^{*}$ ${\ displaystyle V ^ {*}}$ $V^*$ according to the formula

f^{*}(p)=\sup _{x\in M}(\left\langle p,x\right\rangle -f(x)),~~p\in M^{*}=\left\{p:\sup _{x\in M}(\left\langle p,x\right\rangle -f(x))<\infty \right\},

{\ displaystyle f ^ {*} (p) = \ sup _ {x \ in M} (\ left \ langle p, x \ right \ rangle -f (x)), ~~ p \ in M ​​^ {*} = \ left \ {p: \ sup _ {x \ in M} (\ left \ langle p, x \ right \ rangle -f (x)) <\ infty \ right \},}

f^{*}(p)=\sup _{x\in M}(\left\langle p,x\right\rangle -f(x)),~~p\in M^{*}=\left\{p:\sup _{x\in M}(\left\langle p,x\right\rangle -f(x))<\infty \right\},

Where $\left\langle p,x\right\rangle$ ${\ displaystyle \ left \ langle p, x \ right \ rangle}$ Is the value of the linear functional $p$ ${\ displaystyle p}$ on vector $x$ ${\ displaystyle x}$ . In the case of a Hilbert space $\left\langle p,x\right\rangle$ ${\ displaystyle \ left \ langle p, x \ right \ rangle}$ Is an ordinary scalar product. In the particular case of a differentiable function defined in ${\mathcal {R}}^{n}$ ${\ displaystyle {\ mathcal {R}} ^ {n}}$ , the transition to the conjugate function is carried out according to the formulas

f^{*}(p)=\left\langle p,x\right\rangle -f(x),~~~p={\frac {\partial f}{\partial x}}=\operatorname {grad} f,

{\ displaystyle f ^ {*} (p) = \ left \ langle p, x \ right \ rangle -f (x), ~~~ p = {\ frac {\ partial f} {\ partial x}} = \ operatorname {grad} f,}

moreover $x$ ${\ displaystyle x}$ need to be expressed through $p$ ${\ displaystyle p}$ from the second equation.

Geometric meaning

For convex function $f(x)$ ${\ displaystyle f (x)}$ her chart $e p i$ ${\ displaystyle epi}$ $\varphi =\{y|y\geq f(x)\}$ ${\ displaystyle \ varphi = \ {y | y \ geq f (x) \}}$ there is a convex closed set whose boundary is a graph of the function $f(x)$ ${\ displaystyle f (x)}$ . The set of supporting hyperplanes to the function epigraph $f(x)$ ${\ displaystyle f (x)}$ there is a natural domain of definition by its Legendre transformation $f^{*}(p).$ ${\ displaystyle f ^ {*} (p).}$ If a $p$ ${\ displaystyle p}$ - reference hyperplane (in our case, tangent) to the epigraph, it intersects the axis $y$ ${\ displaystyle y}$ at some single point. Her $y$ ${\ displaystyle y}$ -coordinate taken with a minus sign is the value of the function $f^{*}(p)$ ${\ displaystyle f ^ {*} (p)}$ .

Conformity $x\to p$ ${\ displaystyle x \ to p}$ defined uniquely in the region where the function $f(x)$ ${\ displaystyle f (x)}$ differentiable . Then $p$ ${\ displaystyle p}$ - tangent hyperplane to the graph $f(x)$ ${\ displaystyle f (x)}$ at the point $x$ ${\ displaystyle x}$ . Reverse match $p\to x$ ${\ displaystyle p \ to x}$ defined uniquely if and only if the function $f(x)$ ${\ displaystyle f (x)}$ strictly convex. In this case $x$ ${\ displaystyle x}$ - the only point of contact of the reference hyperplane $p$ ${\ displaystyle p}$ with function graph $f(x).$ ${\ displaystyle f (x).}$

If the function $f(x)$ ${\ displaystyle f (x)}$ differentiable and strictly convex, the correspondence is defined $p(x)\leftrightarrow df(x),$ ${\ displaystyle p (x) \ leftrightarrow df (x),}$ matching hyperplanes $p$ ${\ displaystyle p}$ function differential $f(x)$ ${\ displaystyle f (x)}$ at the point $x$ ${\ displaystyle x}$ . This correspondence is one-to-one and allows one to transfer the domain of definition of the function $f^{*}(p)$ ${\ displaystyle f ^ {*} (p)}$ into the space of covectors $V^{*},$ ${\ displaystyle V ^ {*},}$ which are function differentials $f(x)$ ${\ displaystyle f (x)}$ .

In the general case of an arbitrary non-convex function, the geometric meaning of the Legendre transform is preserved. By the support principle, the convex hull of the epigraph $\varphi$ ${\ displaystyle \ varphi}$ is the intersection of half-spaces defined by all supporting hyperplanes; therefore, for the Legendre transform, only the convex hull of the overgraph is essential $\varphi$ ${\ displaystyle \ varphi}$ . Thus, the case of an arbitrary function easily reduces to the case of convex. A function does not even have to be differentiable or continuous; its Legendre transformation will still be a convex lower semicontinuous function.

Properties

Fenchel-Moreau theorem : for a proper convex lower semicontinuous function f defined on a reflexive space, the Legendre transform is involutive , i.e. $f^{**}(x)=f(x)$ ${\ displaystyle f ^ {**} (x) = f (x)}$ . It is easy to see that if the convex closure of f is g , then f * = g *. It follows that for a non-convex function whose convex closure is an eigenfunction,
$f^{**}(x)={\overline {\operatorname {co} }}f(x)$ ${\ displaystyle f ^ {**} (x) = {\ overline {\ operatorname {co}}} f (x)}$ ,
Where ${\overline {\operatorname {co} }}f$ ${\ displaystyle {\ overline {\ operatorname {co}}} f}$ Is the convex closure of the function f .
Directly from the analytical definition follows the Young - Fenchel inequality :
$f(x)+f^{*}(p)\geq \left\langle p,x\right\rangle$ ${\ displaystyle f (x) + f ^ {*} (p) \ geq \ left \ langle p, x \ right \ rangle}$ , and equality is achieved only if p = F '(x).
(Often , Young's inequality is a special case of this inequality for the function F (x) = $x^{a}/a$ ${\ displaystyle x ^ {a} / a}$ , a> 1).
In the calculus of variations (and the Lagrangian mechanics based on it), the Legendre transformation is usually applied to the action Lagrangians $L(t,x,{\dot {x}})$ ${\ displaystyle L (t, x, {\ dot {x}})}$ by variable ${\dot {x}}$ ${\ displaystyle {\ dot {x}}}$ . The Hamiltonian of the action H (t, x, p) becomes the image of the Lagrangian, and the Euler-Lagrange equations for optimal trajectories are transformed into Hamilton equations.
Using the fact that $p=\nabla _{x}f$ ${\ displaystyle p = \ nabla _ {x} f}$ , it is easy to show that $\nabla _{p}f^{*}(p)=-x$ ${\ displaystyle \ nabla _ {p} f ^ {*} (p) = - x}$

Examples

Power Function

Consider the Legendre transform of a function $f(x)=x^{n}$ ${\ displaystyle f (x) = x ^ {n}}$ , ( $n>0$ ${\ displaystyle n> 0}$ , $n\neq 1$ ${\ displaystyle n \ neq 1}$ ) defined on $\mathbb {R^{+}}$ ${\ displaystyle \ mathbb {R ^ {+}}}$ . In the case of even n, we can consider $\mathbb {R}$ ${\ displaystyle \ mathbb {R}}$ .

p(x)={\frac {df}{dx}}=n\cdot x^{n-1}.

{\ displaystyle p (x) = {\ frac {df} {dx}} = n \ cdot x ^ {n-1}.}

From here we express $x=x(p)$ ${\ displaystyle x = x (p)}$ we get

x(p)=\left({\frac {p}{n}}\right)^{\frac {1}{n-1}}.

{\ displaystyle x (p) = \ left ({\ frac {p} {n}} \ right) ^ {\ frac {1} {n-1}}.}

Total we get the Legendre transform for a power function:

f^{*}(p)=px-f(x)=\left({\frac {p}{n}}\right)^{\frac {n}{n-1}}\cdot (n-1).

{\ displaystyle f ^ {*} (p) = px-f (x) = \ left ({\ frac {p} {n}} \ right) ^ {\ frac {n} {n-1}} \ cdot (n-1).}

It is easy to verify that the repeated Legendre transform gives the original function $f(x)$ ${\ displaystyle f (x)}$ .

The function of many variables

Consider the function of many variables defined on the space $\mathbb {R} ^{n}$ ${\ displaystyle \ mathbb {R} ^ {n}}$ of the following form:

f(x)=\langle x,Ax\rangle +c.

{\ displaystyle f (x) = \ langle x, Ax \ rangle + c.}

$A$ ${\ displaystyle A}$ real, positive definite matrix, $c$ ${\ displaystyle c}$ constant. First of all, we verify that the conjugate space on which the Legendre transform is defined coincides with $\mathbb {R} ^{n}$ ${\ displaystyle \ mathbb {R} ^ {n}}$ . To do this, we need to verify the existence of an extremum of the function $\phi =\langle p,x\rangle -\langle x,Ax\rangle -c$ ${\ displaystyle \ phi = \ langle p, x \ rangle - \ langle x, Ax \ rangle -c}$

\nabla _{x}\phi =p-2Ax,

{\ displaystyle \ nabla _ {x} \ phi = p-2Ax,}

\nabla _{x}\nabla _{x}\phi =-2A.

{\ displaystyle \ nabla _ {x} \ nabla _ {x} \ phi = -2A.}

Due to the positive definiteness of the matrix $A$ ${\ displaystyle A}$ , we get that the extremum point is the maximum. So for everyone $p$ ${\ displaystyle p}$ there is a supremum. The calculation of the Legendre transform is carried out directly:

f^{*}(p)={\frac {1}{4}}\langle p,A^{-1}p\rangle -c.

{\ displaystyle f ^ {*} (p) = {\ frac {1} {4}} \ langle p, A ^ {- 1} p \ rangle -c.}

Applications

Hamiltonian mechanics.

In Lagrangian mechanics, a system is described by the Lagrange function. For a typical task, the Lagrange function is as follows:

L(q,u)={\frac {1}{2}}\langle u,Mu\rangle -V(q),

{\ displaystyle L (q, u) = {\ frac {1} {2}} \ langle u, Mu \ rangle -V (q),}

$(q,u)\in \mathbb {R} ^{n}\times \mathbb {R} ^{n}$ ${\ displaystyle (q, u) \ in \ mathbb {R} ^ {n} \ times \ mathbb {R} ^ {n}}$ , with standard, Euclidean scalar product. Matrix $M$ ${\ displaystyle M}$ considered valid, positive definite. In the case when the Lagrangian is not degenerate in velocity, i.e.

p=\nabla _{u}L(q,u)\neq 0,

{\ displaystyle p = \ nabla _ {u} L (q, u) \ neq 0,}

we can make the Legendre transformation in velocities and get a new function called the Hamiltonian:

H(p,q)=pq'-L={\frac {1}{2}}\langle p,M^{-1}p\rangle +V(q).

{\ displaystyle H (p, q) = pq'-L = {\ frac {1} {2}} \ langle p, M ^ {- 1} p \ rangle + V (q).}

Thermodynamics

In thermodynamics, very different thermodynamic functions are very common, the differential of which in the most general case looks like:

dL=Xdx+Ydy+Zdz+...

{\ displaystyle dL = Xdx + Ydy + Zdz + ...}

For example, the differential for internal energy is as follows:

dE=TdS-PdV.

{\ displaystyle dE = TdS-PdV.}

Energy is presented here as a function of variables $S$ ${\ displaystyle S}$ , $V$ ${\ displaystyle V}$ . Such variables are called natural. For example, free energy is obtained as Legendre transformations of internal energy:

F=E-TS,

{\ displaystyle F = E-TS,}

dF=-SdT-PdV.

{\ displaystyle dF = -SdT-PdV.}

In general, if we want to move from a function $L=L(x,y,z,...)$ ${\ displaystyle L = L (x, y, z, ...)}$ to function $L=L(X,y,z,...)$ ${\ displaystyle L = L (X, y, z, ...)}$ , then you should do the Legendre transformation:

L(X,y,z,...)=L-xX,

{\ displaystyle L (X, y, z, ...) = L-xX,}

dL(X,y,z,...)=-xdX+Ydy+Zdz+...

{\ displaystyle dL (X, y, z, ...) = - xdX + Ydy + Zdz + ...}

Field theory. Functional Legendre Transformation.

In quantum field theory, the Legendre functional transformation is very often used. The initial object is the connected Green functions , which are denoted by $W(A)$ ${\ displaystyle W (A)}$ where $A$ ${\ displaystyle A}$ - some external fields. The Legendre transformation over field A is the following function ^[1] :

\Gamma {(\alpha )}=W(A(\alpha ))-\int {dx\cdot \alpha A}.

{\ displaystyle \ Gamma {(\ alpha)} = W (A (\ alpha)) - \ int {dx \ cdot \ alpha A}.}

The sign of integration is usually not written. $\alpha$ ${\ displaystyle \ alpha}$ defined by the following expression ^[1] :

\alpha (x)={\frac {\delta {W}}{\delta {A(x)}}},

{\ displaystyle \ alpha (x) = {\ frac {\ delta {W}} {\ delta {A (x)}}},}

$\delta$ ${\ displaystyle \ delta}$ means the variational derivative . Using the property of the variational derivative, it is easy to derive the following relation $W$ ${\ displaystyle W}$ and $\Gamma$ ${\ displaystyle \ Gamma}$ . Really:

\delta (x-y)={\frac {\delta {A(x)}}{\delta {A(y)}}}=\int {dz{\frac {\delta {A(x)}}{\delta {\alpha (z)}}}{\frac {\delta {\alpha (z)}}{\delta {A(y)}}}}=-\int {dz{\frac {\delta ^{2}\Gamma }{\delta \alpha (x)\delta \alpha (z)}}{\frac {\delta ^{2}W}{\delta A(z)\delta A(y)}}}.

{\ displaystyle \ delta (xy) = {\ frac {\ delta {A (x)}} {\ delta {A (y)}}} = \ int {dz {\ frac {\ delta {A (x)} } {\ delta {\ alpha (z)}}} {\ frac {\ delta {\ alpha (z)}} {\ delta {A (y)}}} = - \ int {dz {\ frac {\ delta ^ {2} \ Gamma} {\ delta \ alpha (x) \ delta \ alpha (z)}} {\ frac {\ delta ^ {2} W} {\ delta A (z) \ delta A (y) }}}.}

In other words, functionals $W_{2}={\frac {\delta ^{2}W}{\delta A(z)\delta A(y)}}$ ${\ displaystyle W_ {2} = {\ frac {\ delta ^ {2} W} {\ delta A (z) \ delta A (y)}}}$ and $\Gamma _{2}={\frac {\delta ^{2}\Gamma }{\delta \alpha (x)\delta \alpha (z)}}$ ${\ displaystyle \ Gamma _ {2} = {\ frac {\ delta ^ {2} \ Gamma} {\ delta \ alpha (x) \ delta \ alpha (z)}}}$ , up to a sign, are inverse to each other. Symbolically, this is written as follows:

W_{2}\cdot \Gamma _{2}=-1.

{\ displaystyle W_ {2} \ cdot \ Gamma _ {2} = - 1.}

Notes

↑ ¹ ² Vasiliev A.N. Functional methods in quantum field theory and statistics. - Leningrad, 1976 .-- S. 81 .-- 295 p.

Literature

Polovinkin E. S, Balashov M. V. Elements of convex and strongly convex analysis. - M .: FIZMATLIT, 2004 .-- 416 p. - ISBN 5-9221-0499-3 .
Vasiliev A. N. Functional methods of quantum field theory and statistics. " —1976. - 295 p.

[:0-1] ¹ ² Vasiliev A.N. Functional methods in quantum field theory and statistics. - Leningrad, 1976 .-- S. 81 .-- 295 p.