# Accumulation point in a pre-image

Let $f:U\subset\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a differentiable function over an open set $U$. If there is a $b\in\mathbb{R}^n$ such that the set $f^{-1}(b)$ has an accumulation point $a\in\mathbb{R}^n$, then $(Df)_a:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is not injective.

-

## 1 Answer

First, your hypothesis assures us that there exists a sequence $a_n \to a,\ a_n \neq a$ with $f(a_n)=b$. By continuity of $f$ this implies that $f(a)=b$. Denote $L=(Df)_a$. Then we know by definition that $L$ is bounded and $$\lim_{h \to 0} \frac{\|f(a+h)-f(a)-L(h) \|}{\|h\|}=0$$

Take $h_n=a_n-a \to 0$ and notice that the above limit gives us $$\lim_{n \to \infty}\frac{L(h_n)}{\|h_n\|}=0 \ \ \ (\star)$$

Suppose that $L$ is injective. Then $L : \Bbb{R}^m \to L(\Bbb{R}^m)$ is bijective and bounded. Because finite dimensional spaces are Banach spaces, it follows that the inverse mapping $$L^{-1} : L(\Bbb{R}^m) \to \Bbb{R}^m$$ is also bounded. This means that there exists a constant $M>0$ ($M$ cannot be zero, since $L^{-1}$ is not the zero operator) such that $\|L^{-1}y\| \leq M\|y\|$ for every $y \in L(\Bbb{R}^m)$. Take $y=Lx$ and get $\|x\| \leq M \|Lx\|$ for every $x \in \Bbb{R}^m$. This contradicts $(\star)$.

Therefore $L$ is not injective.

-