9
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ Just to clarify it for you: So by "useful" you mean "applied to sufficiently many proofs"? $\endgroup$
    – Yes
    Aug 20, 2017 at 8:38
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 20, 2017 at 8:46
  • 2
    $\begingroup$ 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. $\endgroup$
    – Yes
    Aug 20, 2017 at 9:04
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ian
    Aug 20, 2017 at 15:28
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 20, 2017 at 16:55

3 Answers 3

3
$\begingroup$

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 $(1)$.

$\endgroup$
2
  • $\begingroup$ Interesting. In what context did this problem come up for you? $\endgroup$ Aug 20, 2017 at 21:40
  • 2
    $\begingroup$ @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. $\endgroup$ Aug 20, 2017 at 23:10
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $|\det(A)|=1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .