Lagrange multiplier applied to the Legendre transform

Since the Legendre transform is a stationary point of a function, Lagrange multipliers should be the natural choice for handling constraints. However, there is a problem due to the fact that we are mostly interested in its functional dependence on the transform parameter, and not on its value.


Let f(x¯) be a function of the real n-vector x¯ and g(p¯) its Legendre transform defined by

g(p¯)=p¯.x¯-f(x¯)  pi=fxi (1)

The vector p¯ is a function of x¯, or conversely, x¯ is a function of p¯, so that g(p¯) is a function of p¯ alone. The Legendre transform is also alternatively defined as the maximum of the function p¯.x¯-f(x¯). This maximum is reached for x¯ satifying the second set of equations in (1). Suppose now that the componentsPlanetmathPlanetmath xi are not all independent, but are linked by the constraint:

h(x¯)= 0 (2)

This equation defines one of the components, xα for example, as a function of all the others. Putting this function into (1) would give us the transform g(p¯) as a function of the vector p¯ with n-1 components. It would be nice if we could instead use a Lagrange multiplier k and compute the maximum of the function


with all its n components. But then, how does the constraint (2) on x¯ translateMathworldPlanetmath to p¯? The answer is amazingly simple, as we shall see next.


We are going first to compute the Legendre transform the hard way, without the help of a multiplier, by considering xα as a function of the other components:

pi=dfdxi=fxi+fxαxαxi  dhdxi=hxi+hxαxαxi=0  iα (3)

Taking the value of xαxi from the second set of equations (3) and putting it into the first set, we get:

pi=fxi-ΦhxiwithΦ=fxαhxαandiα (4)

From definition (1), the Legendre transform g is therefore:

g(p¯)=iαxi(fxi-Φhxi)-f(x¯) (5)


In the first equation (4), if we set i=α we get pα=0 and conversely, getting the value of Φ. So, in (5), we may remove the condition iα and sum over all the n components of p¯. The constraint pα=0 reduces the number of independent components to n-1. In fact, we are back to the traditional Lagrange method with the multiplier Φ and an additional constraint. We even have the choice between n such constraints. They generate up to n functionally different transforms corresponding to the n possible forms of the function x¯, according to which component xα we choose to eliminate. The method extends easily to the case of more than one constraint:
h1=h2=hm=0. We use m multipliers Φ1,Φ2Φm. They are computed by equating to zero any set of m components from p¯.


  • 1 at PM - Legendre Transform
    This link is actually broken but, hopefully, should be operative soon.
  • 2 - Legendre transformation
Title Lagrange multiplier applied to the Legendre transform
Canonical name LagrangeMultiplierAppliedToTheLegendreTransform
Date of creation 2013-03-22 18:52:47
Last modified on 2013-03-22 18:52:47
Owner dh2718 (16929)
Last modified by dh2718 (16929)
Numerical id 7
Author dh2718 (16929)
Entry type Example
Classification msc 26B10