Use of Lagrange multipliers in pure math problems

Question

I realize that Lagrange multipliers are extremely useful for applied optimization problems.

However, I know that the standard analytic proof of the spectral theorem relies on them. I've also seen a few other uses/mentions of them in some pure math textbooks. (For example, Wade's An Introduction to Analysis uses an exercise on Lagrange multipliers to later prove a result due to Bernstein on the convergence of Fourier series.)

My question, then, is if Lagrange multipliers are generally a useful technique for extremal problems that arise in pure math. If so, are there some well-known proofs in this area that use them (other than those I mentioned)?

I'm simply curious as to their use outside applied optimization, since the derivation of the existence of the so-called Lagrange multiplier is really just a corollary of a very geometric fact--namely that the gradient is perpendicular to the level sets.

EDIT: To be a bit more specific, by "useful" I mean that it is in fact applicable to certain pure math problems with somewhat regular frequency.

Answer 1

Once, I had to solve this problem:

Given two real numbers $a>b>0$, consider the hyperbola$$H=\{(x,y)\in\mathbb{R}^2\,|\,x^2-y^2=a^2-b^2\}.$$One of its points is $P=(-a,-b)$. What is the distance from $P$ to the branch of $H$ to which $P$ does not belong (which, in this case, is the right branch of $H$).

I solved it using Lagrange multipliers. This led me to the system$$\left\{\begin{array}{l}x+a=2\lambda x\\y+b=-2\lambda y\\x^2-y^2=a^2-b^2\\x>0\text{.}\end{array}\right.$$This, in turn, led me to the equation$$\frac{4(a^2-b^2)\lambda^4-3(a^2-b^2)\lambda^2-(a^2+b^2)\lambda}{(1-2\lambda)^2(1+2\lambda)^2}=0\text{.}$$After dividing the numerator by $4(a^2-b^2)\lambda$ (the solution $\lambda=0$ is irrelevant here), one gets a third degree polynomial:$$\lambda^3-\frac34\lambda-\frac{a^2+b^2}{4(a^2-b^2)}.$$Since there is no second degree term, Cardano's formula can be applied directly, giving$$\lambda=\frac12\left(\sqrt[3]{\frac{a-b}{a+b}}+\sqrt[3]{ \frac{a+b}{a-b}}\right)\tag1$$and therefore the point of the right branch of the hyperbola closest to $P$ is $\bigl(-\frac a{1-2\lambda},-\frac b{1+2\lambda}\bigr)$, where the value of $\lambda$ is the one given by $(1)$.

Answer 2

I hope the following example is "pure enough".

Assume $f(x)$ is continuous on $\mathbb{R}$, nonnegative and satisfies $$\int_{-\infty}^{\infty} f(x) dx = 1$$ If $[a,b]$ is an interval with shortest length such that $$\int_{a}^b f(x) dx = \frac{1}{2}$$ Prove that $f(a)=f(b)$.

Define $F(x,y) = \int_x^y f(t) dt$, then $F$ is of $C^1$ on $\mathbb{R}^2$. We wish to minimize $y-x$ subject to constrain $F(x,y)=1/2$. Then, $$(-1, 1) = \nabla (y-x) = \lambda \nabla (F(x,y)) = (-f(x), f(y))$$ Thus, if $[a,b]$ is such that $b-a$ is minimized, we have $$\lambda f(a) = \lambda f(b) = 1$$ This implies $\lambda \neq 0$, hence $f(a) = f(b) = 1/\lambda$.