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I will write the setup of the problem, but I don't think all the parts are necessary to answer my question. If you want the reference, this is from Rabinowitz : Minimax Methods in Critical Point Theory p. 50.

Let $E= W_0^{1,2}(\Omega)$. Suppose that $I(u) = \int_{\Omega} P(x,u) dx$, where $P(x,\xi) = \int_0^{\xi}p(x,t)dt$, and $p(x,\xi)$ is continuous, odd in $\xi$.

It can be shown that $I\in C^1(E,\mathbb{R})$ and $I$ is even.

Here's the claim I don't get:

At a critical point $u$ of $I|_{\partial B_1}$, we have $I'(u)\phi - \mu(u,\phi) = 0$, for all $\phi \in E$.

Could you please explain why this is?

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How is $\mu$ related to the rest of the story? – user53153 Jan 27 '13 at 4:07
That's all it says. Apparently the derivative has this form, where $\mu$ is some constant. Note, if we plug in $\phi = u$, using the fact that $u\in \partial B_1$, we see that $\mu = I'(u)u$. I want to know why it has this form initially though. – Euler....IS_ALIVE Jan 27 '13 at 4:24
This illustrates why using round parentheses for inner product is a bad idea. For all I knew, $\mu(u,\phi)$ was some function $\mu$ with two arguments. – user53153 Jan 27 '13 at 4:27
Sorry. I'm just using the same notation as the author chose. – Euler....IS_ALIVE Jan 27 '13 at 4:36

1 Answer 1

This should be familiar from constrained extremum / Lagrange multiplier topics in multivariable calculus. We are maximizing $I$ subject to the constraint $g=0$ where $g(u)=\langle u,u\rangle -1$. At a critical point $u$ the gradient of $I$ must be proportional to the gradient of $g$. Since $g'(u)\phi=2\langle u,\phi\rangle$, we arrive at $I'(u)\phi = \mu \langle u,\phi\rangle$.

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But what does the gradient mean? This is a functional from a Banach space, and the gradient doesn't have meaning there. The derivative is the frechet derivative, but this also doesn't have meaning on the sphere. That is, what does $I'|_{\partial B_1}$ mean exactly? – Euler....IS_ALIVE Jan 27 '13 at 5:35
@Euler....IS_ALIVE I know what $I'(u)$ means: it is the (Frechet, if you want) derivative of $I$, evaluated at the point $u$. There is no need to differentiate the restriction to the sphere, that is why Lagrange multiplier method is useful. – user53153 Jan 27 '13 at 5:42
Ok, I understand now why this has to be true at a critical point, but the author also claims this holds everywhere on the sphere, not just at the critical points. See my other post… – Euler....IS_ALIVE Jan 27 '13 at 18:29
@Euler....IS_ALIVE What is "this" in "this holds"? The formula in your other post is different; it does not have $\mu$ in it. That formula can be taken as the definition of the derivative of $I_{M}$ where $M$ is a submanifold: $(I_M)'$ is the tangential component of $I'$, that is, $I'$ minus its projection onto the normal vector to $M$. – user53153 Jan 27 '13 at 18:58

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