Talk:Legendre transformation/Archive 1

Convex conjugation
Should the section on convex conjugates be moved to a separate page? -- Tobias Bergemann 10:54, 2004 Nov 4 (UTC)

I have moved the section on convex conjugation to its own page. &mdash;Tobias Bergemann 09:30, 23 November 2005 (UTC)

application of Legendre Transform in thermodynamics
The very idea that ENTHALPY H could be something as a Legendre Transform of the internal energy U is without any scientific evidence. The function (U+pV) was introduced by Rankine in 1854, and named ENTHALPY in 1922. Only in 1960 Callen and Tiszla formulated the hypothesis that enthalpy could be viewed as a Legendre Transform : before that enthalpy was a simple energy-variable ("heat-content at constant pressure"). Any-one interested can get a file with more info on this. &mdash;the preceding unsigned comment is by Smannaerts (talk &bull; contribs) 22:28, 4 January 2006 (UTC1)


 * The idea that enthalpy is a Legendre transform of the internal energy is a mathematically provable fact. Callen and Tiszla did not hypothesize, they recognized that this is true. PAR 05:53, 7 November 2006 (UTC)

History
Could somebody authoritative add some information regarding the history of Legendre transforms? shampoo (talk) 08:34, 7 July 2008 (UTC)

Non-standard Legendre transformation?
I see that in some sections of the article, particularly when it speaks about functions of more than one variable (which is the example of thermodynamic potentials and the Hamiltonian), the Legendre transformation is mentioned as "non-standard". Can anyone explain in which sense it is non-standard, or make any reference? — Preceding unsigned comment added by 88.5.170.179 (talk) 09:15, 13 November 2011 (UTC)

Possible clarifications/improvements
What does this mean:
 * $$f(x) + g(y) = \left\langle x,y\right\rangle.$$

?? Specifically, what is the "inner product"-looking thing? &mdash;the preceding unsigned comment is by TobinFricke (talk &bull; contribs) 20:39, 22 July 2005 (UTC1)

I'm new to all this stuff here. To editing wikipedia as well as of Legendre transformation. But I think this article contains much information -- compared with the german, italian or slovenian page. To clarify it, one should perhaps begin with:


 * Given are two open subsets U and V of Rn and two real-valued differentiable functions f and g such that the first derivative Df is a bijection (one-to-one correspondence) U→V and Dg is a bijection V→U.


 * Then, f and g are said to be Legendre transforms of each other if


 * $$Df = \left( Dg \right)^{-1}$$.

This can still be clarified by explaining how Df must be interpreted as a function U→V, if this is necessary: if x = (x1, ..., xn) is in U and y = (y1, ..., yn), then y = Df(x) means:


 * $$y_i = {\partial f \over \partial x_i}$$

In the following lines, I would replace the confusing expression


 * $$f(x) + g(y) = \left\langle x,y\right\rangle.$$

by


 * $$f(x) + g(y) = \sum_{i=1}^{n}x_{i} \cdot y_{i}$$

In the german article it is said that the Legendre transforms is a special case of the "Berührungstransformation". I don't know what this is, it is not explained in further details. But perhaps "Berührungstransformation" means: Only the condition $$Df = \left( Dg \right)^{-1}$$ holds?

I hope, my suggestions are not too confusing.

-- Mathias Michaelis 12:30, 2005 Oct 25 (MET)

The english term for Berührungstransformation'' would be contact transformation or contactomorphism, which denotes mappings between manifolds that preserve contact structures. Okay, that probably is not too helpful... Maybe a quick look at contact geometry or the page about contact manifolds at PlanetMath helps. That being said: be bold! What you described above would certainly increase the value of this entry.'' &mdash;Tobias Bergemann 09:54, 23 November 2005 (UTC)

What happens whenever Legendre transform is zero?..then most of the properties here couldn't e applied and


 * $$Df = \left( Dg \right)^{-1}$$.

would make no sense. --217.130.79.88 12:19, 13 November 2006 (UTC)


 * The Legendre transform only applies to convex functions $$(d^2f/dx^2>0)$$. f(x)=0 is not convex. PAR 17:23, 13 November 2006 (UTC)


 * The last comment is completely wrong! Only weak convexity is required, and it also very important in applications that it is so. For example in thermodynamics phase transitions in correspond to flat parts of graphs of weakly convex potentials, see the introduction by Wightmann to Israel's book on Lattice's gas. See [Inverse of itself] Esagherardo (talk) 08:03, 21 February 2013 (UTC)

There is a relationship between the Fourier transform of phases (functions that take values on the complex unit circle) that involves the Legendre transform, I heard about it but never found a reference where the relation is stated clearly. Does anybody know where to find about it or add a small section on it if the formulation is simple? [June 2008]

Suggestion for clearer lead
Wouldn't it be clearer if the lead started out with the case of a convex differentiable function? Then instead of this sup stuff, which normal mortals don't handle well, you could just say, "The Legendre transform of a differentiable convex function is the integral of the inverse of its derivative, up to an additive constant. For other functions it may be defined as ..." and give the more general definition.

The point of differentiability is to make the first step defined, and the point of convexity is to make the second step defined by ensuring that the derivative is strictly monotone whence it has an inverse (meaning that the converse of the function as a binary relation is itself a function).

This definition makes it pretty obvious that the Legendre transform is an involution.

Is it a theorem that every differentiable convex function has a Legendre transform, i.e. that the inverse of the derivative is integrable? --Vaughan Pratt (talk) 18:33, 15 October 2011 (UTC)

On second thoughts strict convexity only ensures injectivity of the derivative, whose converse is in general only a partial function. So even in the simple one-dimensional case the object being transformed should be specified as f: A &rarr; B where A and B are real intervals, in order to speak of f as a (total) function. --Vaughan Pratt (talk) 13:07, 16 October 2011 (UTC)


 * The differentiable definition only works for strictly convex functions, while it is important, also for applications, that it is defined on weakly convex functions (e.g. to handle phase transitions in thermodynamics). With the sup definition, which is very easy to handle for normal mortals, this is granted, provided that the definition is complemented with the specification of domains. See the worked out example in my answer to [Inverse of itself] here below. Esagherardo (talk) 08:33, 21 February 2013 (UTC)

Fenchel duality
Among other things, Fenchel's theorem, Fenchel's duality theorem and Fenchel duality were redirected to Legendre transformation. However, I don't remember this page ever having any content about Fenchel duality. Did this content somehow get lost in the wiki history of this page? &mdash;Tobias Bergemann 09:35, 23 November 2005 (UTC)


 * Fenchel was the first who gave a mathematically rigorous treatment of L&eacute;gendre transforms of functions of many variables. Since however this topic was already known as L&eacute;gendre transformations, the redirect is correct. Esagherardo (talk) —Preceding undated comment added 08:36, 21 February 2013 (UTC)

Inverse of itself?
For a convex differentiable function, I can believe that the Legendre transformation is its own inverse, but it appears that the transformation might not be its own inverse in some other cases. For instance, for $$f(x) = ax$$, we get $$f*(x) = x (df/dx) - f = ax - ax = 0$$. I don't see how applying the Legendre transformation to $$f*(x) = 0$$ can possibly get us back to $$f(x) = ax$$. Shouldn't this be clarified in the article?

A more physical example would be the Hamiltonian for a free photon:
 * $$H = c\sqrt{p_x^2 + p_y^2 + p_z^2}\,.$$

It's Lagrangian would be
 * $$L = \vec{p} \cdot \vec\nabla H - H = \vec{p} \cdot \vec{p} c^2 / H - H = H - H = 0\,.$$

How can that be restored to the original H via another application of the Legendre transformation? — Q uantling (talk &#124; contribs) 22:03, 3 March 2011 (UTC)

Rgilbarco (talk) 18:59, 27 September 2011 (UTC)


 * No, its L&eacute;gendre transformation is not 0. It is 1. And it is selfdual!
 * OK, I should be the right person for disentangling the mathematicians/physicists issue, since I am a mathematical physicists! The point is that the definition is incomplete. One has to specify the convex domain where the function is defined; the L&eacute;gendre transform does not depend only on the function, but also on the shape of the domain. The L&eacute;gendre transform of $$f(x)=x$$ is different if it is defined on the line or on an interval. This should be made clear in the definition. Also, much emphasis is put on the differentiable case, which is misleading; although it is true that most important examples are differentiable, it misleads the reader to do wrong computations.
 * OK, I should be the right person for disentangling the mathematicians/physicists issue, since I am a mathematical physicists! The point is that the definition is incomplete. One has to specify the convex domain where the function is defined; the L&eacute;gendre transform does not depend only on the function, but also on the shape of the domain. The L&eacute;gendre transform of $$f(x)=x$$ is different if it is defined on the line or on an interval. This should be made clear in the definition. Also, much emphasis is put on the differentiable case, which is misleading; although it is true that most important examples are differentiable, it misleads the reader to do wrong computations.


 * If $$f:X\mapsto \mathbb R$$ is defined on the convex set $$X$$ and convex, then $$f^*$$ is defined on $$X^*$$ which is the set of values of :$$p$$ for which $$px-f(x)$$ is upper-bounded on $$X$$.


 * This makes a lot of a difference. Take for example $$f(x)=x$$ on $$X=\mathbb R$$. Then $$px-x$$ is upper-bounded on $$X$$ if and only if $$p=1$$, so $$f^*=1$$, but you must record that it is defined on $$X^*=\{1\}$$. Hence it is clear that $$f^{**}=f$$: in fact, $$px-f^*(p)$$ is trivially bounded as a function of $$p$$ on $$X^*$$ for every $$x$$, and its maximum is $$1\cdot x-f^*(1)=x$$. We thus have $$f^{**}(x)=x$$ on $$X^{**}=X=\mathbb R$$. I do not know if such an example should go on a Wikipedia page, but it speaks for making the definition more precise.


 * Note that by construction $$f^*$$ is always continuous and $$X^*$$ is always closed, hence it is impossible that $$f^{**}=f$$ always. For example take the function of $$[-1,1]$$ which is 1 on $$\pm 1$$ and zero in $$ (0,1)$$. It does not fulfil $$f=f^{**}$$. Indeed, convexity implies continuity in open sets, but not on bundaries. The L&eacute;gendre transform is involutive only if the (convex) domain of the function is closed, and the function is continuous on the boundary of its domain.


 * (P.S. It would be nice to correctly spell the name of L&eacute;gendre, which carries an accent.)


 * Esagherardo (talk) 07:37, 21 February 2013 (UTC):

Legendre is spelt without any accent, see for example. — Preceding unsigned comment added by 82.230.69.164 (talk) 16:26, 10 March 2013 (UTC)


 * Ops... apparently you are right. Sorry, and thanks for undoing. Esagherardo (talk) 19:04, 19 March 2013 (UTC)

Misleading picture?
I think the picture is a bit misleading. The derivative of the concave up function f(x) should show up in the negative region of the graph while f(x) is decreasing, no? Without that being shown in the picture, it is a bit misleading. If anyone else agrees, I'll try to post a different picture. Sean Egan (talk) 21:10, 10 October 2013 (UTC)


 * I can't imagine what bothers you, but the picture is 100% correct. If you had your better picture posted here in the talk page, maybe we could discuss it? The picture chooses one arbitrary positive p and produces f*(p) for it. I don't see why a negative slope y=px would help you much. Note f*(p) itself is not plotted: it is the positive segment between the two parallels intersecting the y-axis. Maybe you want to try your idea on Convex conjugate first.  Maybe you want to choose f(x)=exp x, instead, as exemplified later in the article? But, "first do no damage"..... Cuzkatzimhut (talk) 00:09, 11 October 2013 (UTC)

Archive the talk page?
The talk page has posts from almost a decade ago. Would anyone be opposed to archiving old posts, say, created before 2010? Sean Egan (talk) 21:11, 10 October 2013 (UTC)
 * What's the harm? They are so few. Why invite newcomers to rediscover older confusions for themselves? Cuzkatzimhut (talk) 00:11, 11 October 2013 (UTC)

Behaviour under linear transformations
It says


 * Let A be a linear transformation from Rn to Rm. For any convex function f on Rn, one has
 * $$ \left(A f\right)^\star = f^\star A^\star $$
 * where A* is the adjoint operator of A

I don't know what $$A f$$ means: isn't $$ f$$ a function $$\mathbf{R}^n\to \mathbf{R}$$? The result even seems to contradict the scaling result if we think of $$ a$$ as a linear operator on $$\mathbf{R}$$. SimonWillerton (talk) 15:34, 14 November 2013 (UTC)

Ah! The definition of $$A f$$ is on the convex conjugate page. It's the push-forward of $$f$$ along $$A$$.
 * $$ (A f)(y) = \inf\{ f(x) : x \in X, A x = y \} $$

I will add to the page. SimonWillerton (talk) 17:24, 14 November 2013 (UTC)

Abrupt beginning
The abrupt way this article begins with no context-setting and no clear definition should grate upon the sensibilities of thoughtful people. Those who know this topic, please de-stubbify. Michael Hardy 22:13, 4 Feb 2004 (UTC)


 * Yes! Wikipedia is not just for mathematicians. The first sentence really is a joke, with som many references to other concepts. Rbakels (talk) 21:47, 19 April 2014 (UTC)

Problems with the definition of the sets I* and X*
The set


 * $$I^*= \left \{x^*:\sup_{x\in I}(x^*x-f(x))<\infty \right \}$$

is not well defined, as it is nowhere stated if $I* ⊂ R$ or $I* ⊂ C$ or even $I* ⊂ N$. Of course, $X*$ has the same problem:


 * $$X^*= \left \{x^*:\sup_{x\in X}(\langle x^*,x\rangle-f(x))<\infty \right \}$$

A fix would be for example:


 * $$I^*= \left \{x^* \in \mathbb{C} :\sup_{x\in I}(x^*x-f(x))<\infty \right \}$$

--138.246.2.192 (talk) 13:56, 30 June 2014 (UTC)

Huh?
This is absolutely useless as explanation of Legendre transformation. It's not even a definition, let alone explanation.


 * I'm inclined to agree. I don't know the subject, but I know enough to know that whoever wrote this did not write clearly.   Physicists often seem opposed to writing clearly about mathematics; they prefer a touchy-feely style.   Could the author of the above identify "themself"?  Thanks.  Michael Hardy 22:14, 10 Sep 2004 (UTC)


 * Mathematicians often write like Bourbaki and write incomprehensible abstract stuff at a far higher level of abstraction and generality than needed for practical applications and are ever ready to criticize physicists for being "nonrigorous" and "intuitive". &mdash;the preceding unsigned comment is by Tweet Tweet (talk &bull; contribs) 06:53, 12 November 2004 (UTC1)


 *  Mathematicians sometimes write in more generality than is needed for a particular application, but greater abstraction can help make many things clearer and illuminate similarities between different subjects (e.g. the Hamiltonian's use in both classical and quantum mechanics). If mathematicians claim that physicists are not rigorous, it is only because they make statements that are not necessarily clear and obvious or not necessarily true.  That said, there is usually compelling reason to give both clear and precise mathematical definitions and explanations/examples to illustrate what the definitions mean.  75.3.116.185 06:15, 12 June 2007 (UTC)


 * I have a degree in physics, a phd in maths, I teach mathematical physics, and I never have terrible discussions with myself! I think that the definition is a bit poor, and puts too emphasis on the differentiable case. As a clarifying example, see below, the issue [Inverse of itself], and my answer. Esagherardo (talk) 07:51, 21 February 2013 (UTC)

The discussion above not withstanding, the definition given in the main article is no good. It is not exemplified well, and not explained. I have read the entire article twice and did not understand it, then I went and read the definitions by wolfram math and in other place - all of them were easily understandable and non of them had any clear connection to the definition given here. Since I made myself a Wikipedia crush course on C* algebra, and on Riemannian pseudo manifolds, and have had no problem in understanding all the relevant complicated definitions, it is quite obvious to me that this article does not stand up to Wikipedia's standard of clarity, and it should be revised. Note also that Le-genders transformation is a simple tool and not everyone who uses it or look it up in Wikipedia is an expert. — Preceding unsigned comment added by 77.126.161.72 (talk) 09:53, 3 September 2014 (UTC)


 * Please sign your posts. You might try starting from section 4, and working out the examples of section 3.  Is the Zia et al article cited better for you? What exactly are you proposing beyond making faces and talking about yourself? Cuzkatzimhut (talk) 11:33, 3 September 2014 (UTC)

Animation
In case someone likes an animation: http://www.annevanrossum.com/blog/2015/08/08/legendre-transform/ Anne van Rossum (talk) — Preceding undated comment added 21:39, 8 August 2015 (UTC)

`Equivalent definition' is not equivalent.
In the sections `Equivalent definition in the differentiable case' and `Geometric interpretation' it is implied that two smooth functions f and g are Legendre transforms of one another if and only if f' and g' are inverses with respect to composition. It is true that (f*)' and f' are functional inverses. However, here is a counterexample to the other direction of the `equivalence': Define $$ g(p) = f^*(p) + 1 $$ for each p. Then the same as argument as that given in the article shows that g' and f' are functional inverses. Thus, according to the `equivalent definition', the function g is the Legendre transform of f. But g does not equal f*. A constant of integration is needed to fix the choice of Legendre transform among all of its `vertical translates'. For example, one could repair the `equivalent definition' by requiring that g(0)= - f(x').

Lost-n-translation (talk) 19:04, 31 July 2018 (UTC)


 * Ok, I am going to go ahead and repair this section Lost-n-translation (talk) 15:02, 4 August 2018 (UTC)

"Behavior under inversion" formula wrong in general, need reference
It seems to me that the Behavior under inversion formula
 * $$ f(x) = g^{-1}(x) \Rightarrow f^\star(p) = - p \cdot g^\star\left(\frac{1}{p} \right) $$

is wrong in general. I believe it works only for p<0, and is only useful when f is strictly decreasing (and convex).

Also, a reference would be useful (I could not find any). — Preceding unsigned comment added by 84.99.198.15 (talk) 12:22, 3 April 2020 (UTC)

for clarity
I have made this edit,

which generates this text:


 * For sufficiently smooth functions on the real line, the Legendre transform $f*$ of a function $f$ 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
 * $$Df(\cdot) = \left( D f^* \right)^{-1}(\cdot)~,$$ where $$(\phi)^{-1}(\cdot)$$ means a function such that $$(\phi) ^{-1}(\phi(x))=x~,$$

My reason is that, as a visitor, perhaps a dumb visitor, to the article, I wasn't immediately sure that $$( D f^* )^{-1}~,$$ didn't mean $$\frac{1}{D f^*}~.$$

That might be obvious to someone familiar with the topic, but, for a visitor, it isn't immediately clarified by the link to Euler's derivative notation.

Probably a local editor can do a better job of clarification than mine.Chjoaygame (talk) 17:46, 13 January 2021 (UTC)

Example 1 highly misleading
Example 1 is highly misleading. The plot shows the exponential function and its Legendre function both plotted only over the positive reals which suggests that the exponential function also is considered only over the positive reals. In fact, the exponential function is considered over its entire real domain. This is just assumed implicitly. — Preceding unsigned comment added by 87.163.193.23 (talk) 15:52, 23 January 2022 (UTC)

enthalpy defined or derived
This edit https://en.wikipedia.org/w/index.php?title=Legendre_transformation&diff=1167198763&oldid=1167195513 derives enthalpy through a Legendre transform.

As far as I understand, in current 'official' thermodynamics, enthalpy is conventionally defined by the formula $$H = U + PV$$, where $$U$$ is a specifically thermodynamic state variable, defined in terms of a definite state of a working body, of definite composition by mass, and $$P$$ and $$V$$ are ordinary physical variables which contribute to the definition of that state. Such a body, as a thermodynamic system, has a temperature, but this does not conventionally enter into the present definition of its thermodynamic state. In this definition, it seems to me inappropriate to define $$P$$ as $$\partial U / \partial V.$$ That needs to be derived from the definition.

So I ask, should one show that the conventional definition of the enthalpy leads to the Legendre transform statement about it, rather than showing that the Legendre transform statement leads to the conventional definition?Chjoaygame (talk) 09:40, 8 August 2023 (UTC)

For a logical development, I suppose one would start with a reference state of a closed working body, using an empirical thermometer to fix a reference temperature $$T_0$$. One would then define $$U_{T_0} = U_{T_0}(V)$$ through $$\Delta U = -W = \int_A^B P \  \text{d}V$$, with empirical measurements of $$P$$ and $$V$$ as ordinary physical variables, using slow adiabatic expansions of the working body starting or ending at $$T_0$$. One could then introduce thermodynamic temperature (and thence entropy) by calorimetry through a reference internal energy $$U_{T_0}$$ along with $$\Delta U = Q$$ and $$\text{d}S = \text{đ}Q / T$$ at suitable constant volumes. One would then have two formulas for $$U,$$ namely $$U_{T} = U_{T}(P,V)$$ and $$U = U(S,V).$$Chjoaygame (talk) 11:40, 11 August 2023 (UTC)