The n-Category Café

From Legendre to Laplace: the rough idea

In Part 1 we saw that for each $\beta \gt 0$ we can make $(0,\infty]$ into a rig with addition

$$x \oplus_\beta y = -\frac{1}{\beta} \ln(e^{-\beta x} + e^{-\beta y})$$

As $\beta \to +\infty$, this operation approaches

$$x \oplus_\infty y = x \min y$$

and we get a version of the tropical rig. (There’s a more popular version using $\max$ instead of $\min$, but I decided to use $\min$.)

Suppose we want to define a notion of integration with $\oplus_\beta$ replacing ordinary addition. We could call it ‘$\beta$-integration’ and denote it by $\int_\beta$. Then it’s natural to try this:

$$\int_\beta f(x) \, d x = - \frac{1}{\beta} \ln \int e^{-\beta f(x)} d x$$

If the function $f$ is nice enough, we could hope that

$$\lim_{\beta \to +\infty} \int_\beta f(x) \, d x \; = \; \inf_x f(x)$$

Then, we could hope to express the Legendre transform

$$\tilde{f}(s) = \inf_{x \in \mathbb{R}} (s x - f(x))$$

as the $\beta \to + \infty$ limit of some transform involving $\beta$-integration.

Indeed, in Section 6 here, Litvinov claims that the Legendre transform is the $\beta = \infty$ analogue of the Laplace transform:

• Grigory L. Litvinov, Tropical mathematics, idempotent analysis, classical mechanics and geometry.

But he doesn’t state any result saying that it’s a limit of the Laplace transform, or something like that.. Touchette states a result along these lines here:

• Hugo Touchette, Legendre–Fenchel transforms in a nutshell.

He even applies it to classical statistical mechanics! But he’s operating as a physicist, not a mathematician, so he doesn’t state a precise theorem. I’d like to take a crack at that, just to be sure I’m not fooling myself.

The Legendre transform as a limit

Touchette’s formula gives an Legendre transform involving a sup rather than an inf. I slightly prefer a version with an inf. We’ll get the Legendre transform as a limit of something that is not exactly a Laplace transform, but close enough for our physics application:

Almost Proved Theorem. Suppose that $f \colon [0,\infty) \to \mathbb{R}$ is a concave function with continuous second derivative. Suppose that for some $s \gt 0$ the function $s x - f(x)$ has a unique minimum at $x_0$, and $f''(x_0) \lt 0$. Then as $\beta \to +\infty$ we have

$$\displaystyle{ -\frac{1}{\beta} \ln \int_0^\infty e^{-\beta (s x - f(x))} \, d x \quad \longrightarrow \quad \inf_x \left( s x - f(x)\right) }$$

Almost Proof. Laplace’s method should give the asymptotic formula

$$\displaystyle{ \int_0^\infty e^{-\beta (s x - f(x))} \, d x \quad \sim \quad \sqrt{\frac{2\pi}{-\beta f''(x_0)}} \; e^{-\beta(s x_0 - f(x_0)} }$$

Taking the logarithm of both sides and dividing by $-\beta$ we get

$$\displaystyle{\lim_{\beta \to +\infty} -\frac{1}{\beta} \ln \int_0^\infty e^{-\beta(s x - f(x))} \, d x \; = \; s x_0 - f(x_0) }$$

since

$$\displaystyle{ \lim_{\beta \to +\infty} \frac{1}{\beta} \; \ln \sqrt{\frac{2\pi}{-\beta f''(x_0)}} = 0 }$$

Since $s x - f(x)$ has a minimum at $x_0$ we get the desired result:

$$\displaystyle{ -\frac{1}{\beta} \ln \int_0^\infty e^{-\beta (s x - f(x))} \, d x \quad \longrightarrow \quad \inf_x \left( s x - f(x)\right) }$$

The only tricky part is that Laplace’s method as proved here requires finite limits of integration, while we are integrating from $0$ all the way up to $\infty$. However, the function $s x - f(x)$ is concave, with a minimum at $x_0$, and it has positive second derivative there since $f''(x_0) \lt 0$. Thus, it grows at least linearly for large $x$, so as $\beta \to +\infty$ the integral

$$\displaystyle \int_0^\infty e^{-\beta(s x - f(x))} \, d x$$

can be arbitrarily well approximated by an integral over some finite range $[0,a]$.       ∎

Someone must have studied the hell out of this issue somewhere — do you know where?

Now, let’s look at the key quantity in the above result:

$$\displaystyle{ - \frac{1}{\beta} \ln \int_0^\infty e^{-\beta (s x - f(x))} \, d x }$$

We’ve got a $-1/\beta$ and then the logarithm of an integral that is not exactly a Laplace transform… but it’s close! In fact it’s almost the Laplace transform of the function

$$g_\beta(x) = e^{\beta f(x)}$$

since

$$\displaystyle{ \int_0^\infty e^{-\beta (s x - f(x))} \, d x = \int_0^\infty e^{-\beta s x} \, g_\beta(x) \, d x }$$

The right hand side here would be the Laplace transform of $g_\beta$ if it weren’t for that $\beta$ in the exponential.

So, it seems to be an exaggeration to say the Legendre transform is a limit of the Laplace transform. It seems to be the limit of something that’s related to a Laplace transform, but more complicated in a number of curious ways.

This has made my life difficult (and exciting) for the last few weeks. Right now I believe that all the curious complications do exactly what we want them to in our physics applications. But I should hold off on declaring this until I write up all the details: I keep making computational mistakes and fixing them, with a roller-coaster of emotions as things switch between working and not working.