While reading a necessary condition for an extremum with constraint from Zorich's book (volume I, page 528) I asked myself the following question:
Question. Let $D$ be an open subset in $\mathbb{R}^n$ and $f:D\to \mathbb{R}$ such that $f\in C^1(D;\mathbb{R})$. Assume that $x_0\in D$ and $\nabla f(x_0)\neq\vec{0}$. Consider $N:=\{x\in D: f(x)=f(x_0)\}$. How to prove that $N$ is an $(n-1)$-dimensional submanifold of $\mathbb{R}^n$?
Zorich gives the following definition of a $k$-dimensional submanifold of $\mathbb{R}^n$.
Definition 1. We shall call a set $S\subset \mathbb{R}^n$ a $k$-dimensional smooth surface in $\mathbb{R}^n$ (or a $k$-dimensional
submanifold of $\mathbb{R}^n$) if for every point $x_0\in S$ there
exist a neighborhood $U(x_0)$ in $\mathbb{R}^n$ and a diffeomorphism
$\varphi:U(x_0)\to I^n$ of this neighborhood onto the standard
$n$-dimensional cube $I^n=\{t\in \mathbb{R}^n: |t^i|<1,\ i=1,\dots,n\}$
of the space $\mathbb{R}^n$ under which the image of the set $S\cap
U(x_0)$ is the portion of the $k$-dimensional plane in $\mathbb{R}^n$
defined by the relations $t^{k+1}=0,\dots, t^n=0$ lying inside $I^n$.
EDIT 1: I can answer my question if we assume $\nabla f(x)\neq \vec{0}$ for all $x\in D$. I will show that $N$ is a $(n-1)$-dimensional submanifold of $\mathbb{R}^n$. Indeed, let $a=(a_1,\dots,a_n)\in N$ then we know that $\nabla f(a)\neq \vec{0}$. WLOG, assume that $\frac{\partial f}{\partial x^1}(a)\neq 0$. Consider the mapping $\Phi:D\to \mathbb{R}^n$ defined as $\vec{x}\mapsto(f(x)-f(x_0),x^2,\dots,x^n)$, where $x=(x^1,\dots,x^n)$. It is not difficult to see that $\Phi\in C^1(D;\mathbb{R}^n)$ and $\det J_{\Phi}(a)\neq 0$, where $J_{\Phi}(a)$ is a Jacobian of our mapping at $a$. By inverse function theorem there exist $O_1=O_1(a)$ neighborhood of $a$ and $O_2=O_2(\Phi(a))$ neighborhood of $\Phi(a)$ such that $\Phi:O_1\to O_2$ is a diffeomorphism. We notice that $\Phi(a)=(0,a_2,\dots,a_n)$ and we can find a cube around $\Phi(a)$ which is contained in $O_2$ and assume that $$Q=(-r,r)\times \prod_{j=2}^n (a_j-r,a_j+r).$$
Let $P:=\Phi^{-1}(Q)$, then $P\subset O_1$ and $P$ is a neighborhood of $a$ (because $\Phi$ is a diffeomorphism). Therefore, for $a\in N$ we were able to find its neghborhood $P=P(a)$ and diffeomorphism $\Phi$ such that $\Phi:P(a)\to Q$. Moreover, it is not difficult to check that $$\Phi(P(a)\cap N)=\{0\}\times \prod_{j=2}^n (a_j-r,a_j+r).$$ By definition 1, it follows that $N$ is a $(n-1)$-dimensional submanifold.
Remark. But in the original question we assumed that $\nabla f(x_0)\neq 0$. WLOG suppose that $\partial_1 f(x_0)\neq 0$ Since $f\in C^1(D;\mathbb{R})$, then there exists $O(x_0)\subset D$, where $O(x_0)$ is a neighborhood of $x_0$ such that $\partial_1f(x)\neq 0$ for all $x\in O(x_0)$. It implies that $\nabla f(x)\neq \vec{0}$ for all $x\in O(a)$. Using previous reasoning it implies that $\tilde{N}:=\{x\in O(x_0): f(x)=f(x_0)\}$ is a $(n-1)$-dimensional submanifold. But it does not imply that $N$ is a $(n-1)$-dimensional submanifold.
I hope now my question is posed correctly and it maked sense what am I asking.
Thank you!
EDIT 2: I created this post because Zorich proves the following result in his book:
Theorem 1. Let $f:D\to \mathbb{R}$ be a function defined on an open set $D\subset \mathbb{R}^n$ and belonging to $C^1(D;\mathbb{R})$.
Let $S$ be a smooth surface in $D$. A necessary condition for a point
$x_0\in S$ that is noncritical for $f$ to be a local extremum of
$\left.f\right|_S$ is that $$TS_{x_0}\subset TN_{x_0},$$ where
$TS_{x_0}$ is the tangent space to the surface $S$ at $x_0$ and
$TN_{x_0}$ is the tangent space to the level surface $N = \{x \in D: f(x) = f(x_0)\}$ of $f$ to which $x_0$ belongs.
So my concern was that one cannot define surface $N$ in the way he did. I mean $N$ should be defined as $\{x\in O(x_0): f(x)=f(x_0)\}$, where $O(x_0)\subset D$ is a neighborhood of $x_0$, where $\nabla f(x)\neq 0$. Is that correct?
Best Answer
The only technical issue with the Theorem 1 you quote is that the definition of $N$ is over all of $D$. As another answerer has pointed out, this is false in general. However, Theorem 1 is a local criterion (whether or not a point is an extremum) and so really the set $D$ can be restricted (without loss of generality) to a sufficiently small open set containing $x_0$ so that $\nabla f$ is non-zero on $D$ and so $N$ is sufficiently well-defined. Indeed if $f$ is $C^1$ and $\nabla f(x_0)$ is non-zero then there exists a restriction of the open set $D$ containing $x_0$ where $\nabla f$ is non-zero everywhere and a further restriction (if required) to keep $N$ well-defined. Said differently, once $\nabla f(x_0)\neq 0$ is known then we can shrink $D$ without loss of generality to make $N$ well-defined thereby making the necessary condition testable.
It would be overly pedantic (in my opinion) to consider a smaller neighbourhood of $D$ since the theorem is a necessary condition anyway. Being unable to construct $N$ on some set $D$ does not mean no restriction of $D$ exists where you can construct it --- assuming that $x_0$ is a non-critical point of course.