# Basic Constrained functional form

Let $I \in C^1(\mathbb{R}^n,\mathbb{R})$ be an even functional. There is a claim that then the restriction to the sphere has the following form for the Frechet derivative $$I|_{S^{n-1}}'(u) = I'(u)-(I'(u),u)u$$ I don't see why this is true. Could you explain please?

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The Fréchet derivative does not make sense here, because $I|_{\mathbb{S}^{n-1}}$ is not a function defined on an open set of a linear space. I think what is meant here is the gradient of $I|_{\mathbb{S}^{n-1}}$ as a function on the Riemannian manifold $\mathbb{S}^{n-1}$. That is a vector field, i. e. a section of the tangent bundle of the sphere, so embedded in the tangent bundle of $\mathbb{R}^n$, at each point $u \in \mathbb{S}^{n-1}$ it has to be orthogonal to $u$. In fact, it is the orthogonal projection of the gradient of $I$ in $\mathbb{R}^n$ along $u$, and that is what the formula is supposed to mean.
Do you mean, the gradient of I (that is, not the gradient of $I|_{S^{n-1}}$) as a function on $S^{n-1}$, since you just said that the derivative does not make sense here? – Euler....IS_ALIVE Jan 25 '13 at 18:15
also if this expression is orthogonal to $u$, then that means that we should (evaluating at $u$) that $(I'(u),u) - (I'(u),u)u = 0$. This doesn't seem to be the case. – Euler....IS_ALIVE Jan 25 '13 at 22:07
@Euler....IS_ALIVE Frechet derivative (a concept of functional analysis) does not make sense for $I_{|S^{n-1}}$, but the gradient (concept of Riemannian geometry) does. – user53153 Jan 25 '13 at 22:51
@Thomas I would be surprised, since the author makes no reference to Riemannian geometry. This is in fact from a PDE book. The goal is to prove that under the conditions for $I$ above, then $I|_{S^{n-1}}$ has at least $n$ distinct pairs of critical points, apparently first proved by Ljusternik and Schnirelmann. This link is not the book I am studying, but it is by the same author with the same claim tinyurl.com/aq27j92 p.153-154 – Euler....IS_ALIVE Jan 25 '13 at 23:02
@Euler....IS_ALIVE It is orthogonal to $u$: $(I'(u)-(I'(u),u)u,u) = 0$, since $(u,u) = 1$. I guess the author implicitely parametrizes the sphere as a submanifold of $\mathbb{R}^n$, concatenates the parametrization with $I$. Then he can take the gradient, but has to push it forward to $\mathbb{R}^3$ via the derivative of the parametrization. – Thomas Jan 27 '13 at 14:50