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When I read the Wiki about Legendre transformation, there is a statement

The Legendre transformation is an application of the duality relationship between points and lines.

What's the meaning of this statement?

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  • $\begingroup$ Legendre transformations connect point geometry and Pluecker geometry. Check this link as well. $\endgroup$ – user35952 Apr 4 '14 at 8:31
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For a convex function you can do the following:

For each point on the graph of the function, draw the line tangent to the function at that point. That point can now be identified by its original $x$ and $y=f(x)$ coordinates, or by specifying the slope of that tangent line and its corresponding y-intercept. Each point maps to one and only one line, and vice versa. For convex functions, the mapping is one-to-one. There is no ambiguity. Draw a sketch and you will soon be convinced.

The Legendre transformation gives you the value of the y-intercept if you give it the slope. So the Legendre transform is a plot of $b(m)$ vs $m$ (y-intercept as a function of slope) rather than $f(x)$ vs $x$. Either function represents the same data or concept. In a sense they contain the same information.

update Thanks to @EmilioPisanty for improved wording. See comments.

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  • $\begingroup$ This is not quite correct (in language; the content is OK). You can't specify both the slope and the y-intercept of a line, in the same way that you don't have freedom over $x$ and $y$ independently. $\endgroup$ – Emilio Pisanty Apr 4 '14 at 13:56
  • $\begingroup$ If I understand you correctly, I agree completely. Is this it? --- given $x$, then $y=f(x)$ is specified and "fixed". Similarly, given $\mathrm{d}f/\mathrm{d}x$, the L.T. specifies (and "fixes") the y-intercept (of the line having that slope that is also tangent to $f(x)$ somewhere). $\endgroup$ – garyp Apr 4 '14 at 14:08
  • $\begingroup$ I would phrase your third sentence as something like "That point can now be identified by its original $x$ adn $y=f(x)$ coordinates, or by specifying the slope of that tangent line and its corresponding $y$-intercept." (But of course that's my phrasing.) $\endgroup$ – Emilio Pisanty Apr 4 '14 at 14:11
  • $\begingroup$ Yes, it's your phrasing, but I still like it better than mine. Thanks. Editing answer. $\endgroup$ – garyp Apr 4 '14 at 14:13

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