# Use of Lagrange multipliers in pure math problems

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.

• Just to clarify it for you: So by "useful" you mean "applied to sufficiently many proofs"?
– Yes
Aug 20, 2017 at 8:38
• @GaryMoore I left the meaning vague because I'm interested in hearing about any way in which pure mathematicians may find them helpful. But the definition you give probably works well. Aug 20, 2017 at 8:46
• You don't have to explain it for me:). Perhaps I should have said it in advance, That I left the comment is out of my experience here: The more equivocal an asker here sounds, the more likely the question gets closed.
– Yes
Aug 20, 2017 at 9:04
• The Euler-Lagrange equation can be thought of as a kind of Lagrange multiplier condition in functional space. Of course this is at the interface between pure and applied math, as differential equations frequently are.
– Ian
Aug 20, 2017 at 15:28
• Lagrange multipliers can often be used to prove inequalities between numbers. A search engine query with words such as lagrange multipliers hardy littlewood polya yields quite a few results. Aug 20, 2017 at 16:55

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 $$(1)$$.

• Interesting. In what context did this problem come up for you? Aug 20, 2017 at 21:40
• @CuriousKid7 Consider the rectangle whose vertices are $(\pm a,\pm b)$. I wanted to know, given a number $r\in(0,2\sqrt{a^2+b^2})$, how many points within the rectangle can have the intersection between the hyperbola $H$ and the circle centered at $(-a,-b)$ with radius $r$. The answer is $1$, $2$, or $3$, and it can only be $3$ when $r>2b$ and $r$ is smaller than the distance from $(-a,-b)$ to the branch of $H$ that doesn't contain it. So, I was led to the computation of this distance. Aug 20, 2017 at 23:10

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 $f(a) = f(b) = 1/\lambda$.

You can use Lagrange multipliers for the proof of Hadamard's inequality. It says that for a square matrix $$A$$ with column vectors $${\bf a}_k$$ one has $$\left|\det(A)\right|\leq\prod_{k=1}^n|{\bf a}_k|\ .$$ The claim can be written in the form $$-1\leq f({\bf a}_1,\ldots,{\bf a}_n)\leq1\ ,\tag{1}$$ whereby I have considered the determinant $$f$$ as a function of $$n$$ vector variables, and it is assumed that all $$|{\bf a}_k|=1$$. In other words, we have the $$n$$ constraints $$F_k({\bf a}_1,\ldots,{\bf a}_n):={1\over2}\bigl(a_{1k}^2+a_{2k}^2+\ldots+a_{nk}^2-1\bigr)=0\qquad(1\leq k\leq n)\ ,$$ each of them reigning over just $$n$$ of the $$n^2$$ real variables $$a_{ik}$$. Lagrange's principal function then is $$\Phi=f-\sum_{k=1}^n\lambda_k F_k\ ,$$ and we have to differentiate $$\Phi$$ with respect to all variables $$a_{ik}$$. We obtain the $$n^2$$ equations $$\left({\partial \Phi\over\partial a_{ik}}=\right)\quad A_{ik}-\lambda_k a_{ik}=0\qquad({\rm all}\ i, k)\ ,\tag{2}$$ where $$A_{ik}$$ is the cofactor of $$a_{ik}$$ in the determinant $$f$$.

For the rest of the proof some linear algebra is used. Multiply $$(2)$$ by $$a_{ir}$$, and sum over $$i$$: $$\sum_{i=1}^n a_{ir}A_{ik}-\lambda_k\>{\bf a}_r\cdot{\bf a}_k=0\qquad ({\rm all}\ r, k)\ ,$$ which is \lambda_k\>{\bf a}_r\cdot{\bf a}_k=\left\{\eqalign{\det(A)\quad&(r=k)\cr 0\qquad&(r\ne k)\cr}\right.\qquad ({\rm all}\ r, k)\ . Letting $$r=k$$ we see tha all $$\lambda_k$$ have the same value $$\det(A)\ne0$$ in the extremal situation, and letting $$r\ne k$$ we then see that $${\bf a}_r\cdot{\bf a}_k=0$$ in this case. We therefore have $$A'\cdot A=$$ identity matrix, and this implies $$|\det(A)|=1$$.