# The Frechet derivative and Lagrange multipliers on Banach spaces

I am interested in questions of the following form: minimize $H(f)$ given $G(f) = 0$ where $H$ and $G$ are operators of type $X \to R$ where $X = R \to R$. An example is:

Minimize $$H(f) = \int_{-1}^1\sqrt{1+f'(x)^2}$$

Under the conditions that:

$$G(f) = \int_{-1}^1f(x) = \pi/2$$ $$f(-1) = f(1) = 0$$

That is, find a function f on [-1,1] with area under the curve equal to $\pi/2$, minimizing the path length (the answer to this example is a half-circle $f(x) = \sqrt{1-x^2}$, I believe).

Other examples of operators that can occur in the minimize or in the condition are:

$$A(f) = f(1)$$

$$B(f) = \int_a^b E(x,f(x),f'(x))dx$$

(i.e. an integral of an expression containing $x$, $f(x)$ and $f'(x)$).

The way to solve these kind of problems seem to be Lagrange multipliers on Banach spaces. How does one do this?

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The optimum in the first case looks to me like it should be a constant function. – Noah Stein Jan 26 '11 at 16:47
Dear Jules, your post could be a bit more easy on the eyes if you use the LaTeX support on this website. – Willie Wong Jan 26 '11 at 16:49
Hmm yes you are right...how about the problem with the additional constraint f(0) = f(1) = 0? – Jules Jan 26 '11 at 16:49
It depends what you mean by a solution, and the level of rigour you want. Physicists and engineers solve similar problems non-rigorously all the time, by Calculus of Variations methods (which involve Lagrange multipliers). Euler-Lagrange equations are the keywords to search for. However, they usually don't specify exactly which functions they consider. Banach spaces only arise if your restrictions are explicit: $f$ being $C^1$, $C^2$, etc., and you define a complete norm on the class of functions. But for most specific problems, Banach space theory is probably not worth the effort. – Zen Harper Jan 26 '11 at 18:27
It seems to me that the calculation part of Calculus of Variations (i.e., how to find the Euler-Lagrange equations), including constrained problems, is presented in many physics, namely classical mechanics, textbooks. – Deane Yang Jan 26 '11 at 20:33

NOTE: this was a comment, because I thought it wasn't detailed enough for an answer; but Jules (the OP) specifically asked me to post it as an answer.

NOTE to Jules: However, maybe you should wait a few hours or days before accepting any answer, to give others a chance to read it (differing time zones around the world, etc.) [Some MO people seem to get a bit annoyed when people accept answers very quickly].

Sorry I can't think of any particular book to recommend for Calculus of Variations; but I think an advanced undergraduate book (maybe just a chapter or two of a "Mathematical Methods" book) might be more useful than a detailed Graduate level book, which would probably be more than you need.

It depends what you mean by a solution, and the level of rigour you want. Physicists and engineers solve similar problems non-rigorously all the time, by Calculus of Variations methods (which involve Lagrange multipliers). Euler-Lagrange equations are the keywords to search for. However, they usually don't specify exactly which functions they consider. Banach spaces only arise if your restrictions are explicit: being $C^1$, $C^2$, etc., and you define a complete norm on the class of functions. But for most specific problems, Banach space theory is probably not worth the effort.

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Wikipedia has Lagrange multipliers on Banach spaces. Recall that $\mathbb R$ is a Banach space; the original question has it as the codomain of the constraint function. In this case the Lagrange multiplier is a simple multiplier, that is, a linear function $\mathbb R \rightarrow \mathbb R$. I'll write $\lambda(x) : x \mapsto \lambda x$. Also important to keep in mind is that the existence of Langrage multipliers is a necessary condition but not a sufficient one. Therefore calculating a Lagrange multiplier only yields candidate extrema.

In the present example, to satisfy the conditions for the existence of a Langrange multiplier, we need $G(x) = \int f(x) dx - \pi/2$ so that the constraint is of the form $G(x) = 0$. This won't matter in the present case, but it's required in more complicated ones. Thus we have $$DH_f(g) = \int_{-1}^1 \frac{ f'(x) }{ \sqrt{1+ f'(x)^2}} g' dx \qquad DG_f(g) = \int_{-1}^1 g(x) dx$$ A local extremal of $H$ at $f_0$ subject to $G(x)=0$ has a Lagrange multiplier $\lambda$ such that $$DH_{f_0} = \lambda DG_{f_0}$$ Equality is that for functions, so these must be true for all arguments $g$. Note that constant functions $f$ have $DH_f=0$ (since the integrand is zero). Constant functions are candidate local extrema for $H$ with respect to any constraint by choosing multiplier $\lambda=0$. In the present case, however, the constraint set contains no constant functions.

If we didn't know the answer, we could integrate the expression for $DH_f$ by parts and thereby calculate a variational derivative for $H$. Since we have a candidate answer, we can skip that, perform the substitution first, and then integrate by parts. I omit the algebra: $$f_0(x) = \sqrt{1+x^2} \qquad f_0'(x) = \frac{ -x }{ \sqrt{1+x^2} } \\ DH_{f_0} = \int_{-1}^1 (-x) g' dx = \int_{-1}^1 g(x) \, dx$$ Choosing $\lambda=1$, we see that $f_0$ is a candidate extremal of $H$ subject to $G=0$.

I'm done, but a full proof that $f_0$ is a minimum is not. Existence of the Lagrange multiplier, as I said above, is not sufficient. One approach to proving minimality would be as follows:

• Compute the variational derivative $\delta H$, that is, $DH_f(g) = \int \delta H_f(x) \, g \, dx$. The multiplier equation is then $\int (\delta H_f(x) - \lambda \circ \delta G_f(x)) g(x) \, dx$, and we have $\delta G = 1$.
• Show that the operator $g \mapsto \int (\delta H_f(x) - \lambda) g(x) \, dx$ is non-degenerate. Anything in a non-trivial eigenspace of $\delta H$ is a candidate extremal, and this step eliminates them.
• Show that $f_0$ is the unique solution within the constraint set to the differential equation $\delta H_f(x) - \lambda = 0$ for any $\lambda$.
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