Isn't this a trivial corollary?

Question

Let $U \subseteq \mathbb R^{n}$ be an open subset and let $M \subseteq U$ be a $k$-dimensional submanifold of $\mathbb R^{n}$. Consider a differentiable function $f: \ U \longrightarrow \mathbb R$. There's a corollary that states, that if $f \big|_M$ takes on a local extremum at a point $p \in M$, then the gradient $\nabla f(p)$ is normal to $M$ at $p$, i.e. $\nabla f(p) \perp T_{p}M$, where $T_{p}M$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $\nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?

Answer 1

Take$$M=\{(x,y)\in\mathbb R^2\mid x^2+y^2=1\}$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $\nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).

The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $\mathbb R^n$), but points of submanifolds usually are not interior points.

Answer 2

$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.

Consider for instance $f$ defined on $\mathbb{R}^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $\nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^\bot$, but is not $0$