Let $U\subset\mathbb{R}^m$ be an open set and $f:U\to\mathbb{R}^n$ a differentiable function. Suppose that there exists $b\in\mathbb{R}^n$ and $a\in U$ such that $a$ is an accumulation point of $f^{-1}(b)=\{x\in U;\;f(x)=b\}$.

The problem is to show that $f'(a):\mathbb{R}^m\to\mathbb{R}^n$ is not injective.

Since $a$ is an accumulation point of $f^{-1}(b)$, there exists a sequence $(x_k)$ in $f^{-1}(b)$ such that $x_k\neq a$ for all $k\in\mathbb{N}$ and $\lim x_k=a$. By continuity of $f$, it follows that $f(a)=\lim f(x_k)=b$.

We know that, if $t_k\to 0$ in $\mathbb{R}$ and $v_k\to v$ in $\mathbb{R}^m$, then

$$f'(a)\cdot v=\lim \frac{f(a+t_kv_k)-f(a)}{t_k}=\lim_{t\to 0}\frac{f(a+tv)-f(a)}{t}=\frac{\partial f}{\partial v}(a).$$

If $m=n=1$ we can take $t_k=x_k-a$, $v_k=1$ and conclude that

$$f'(a)\cdot 1=\lim \frac{f(a+t_k)-f(a)}{t_k}=\lim \frac{f(x_k)-f(a)}{x_k-a}=\lim\frac{0}{x_k-a}=0.$$

So, $f'(a)$ is not injective because $1\in \operatorname{Ker} f'(a)$.

How can we deal with the general case?



3 Answers 3


First prove the following fact: if $L:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is an injective linear mapping, there exists an $\epsilon>0$ such that for all $x\in \mathbb{R}^n:\left\|L\left(\frac{x}{\|x\|}\right)\right\|\ge\epsilon$.

Proof: suppose such an $\epsilon>0$ doesn't exist. Then we can find a sequence $(x_n)$ with $\|x_n\|=1$ such that $\|Lx_n\|\rightarrow0$. Now choose a convergent subsequence $(x_{n_k})$. Since $\|x_{n_k}\|=1$, we must have $x:=\lim_{k\to \infty}x_{n_k}\neq 0$. But then it follows that $$0=\lim_{k\to \infty}\|Lx_{n_k}\|=\|\lim_{k\to \infty}Lx_{n_k}\|=\|Lx\|.$$

Hence $Lx=0$, which contradicts the injectivity of $L$.

Now, to prove your statement, choose a sequence $(a_n)$ in $f^{-1}(b)$ such that $a_n\rightarrow a$. Put $x_n:=a_n-a$. By the definition of the derivative, we have: $$0=\lim_{n\to \infty}\frac{f(a+x_n)-f(a)-Df(a)x_n}{\|x_n\|}=\lim_{n \to \infty}Df(a)\left(\frac{x_n}{\|x_n\|}\right),$$

since $f(a+x_n)=f(a)$. It follows that $Df(a)$ cannot be injective.

  • $\begingroup$ Here there are other proofs of your first fact. $\endgroup$
    – Pedro
    Jun 27, 2014 at 20:48

We can apply the definition of the derivative in $n$ dimesnions. Let $A=f^{\prime}(a)$ Then this means that

$$\lim\limits_{|\mathbf{h}|\to 0} \frac{|f(a+\mathbf{h})-f(a)-A\mathbf{h} | }{|\mathbf{h}|}=0$$ Now as you point out, $f(a)=b$ and it is possible to chose arbitrarily small $|\mathbf{h}|$ such that $f(a+\mathbf{h})=b$ and so what we have is that for every $\epsilon >0$ there is a non zero $\mathbf{h}$ such that $$|A\mathbf{h}|\leq\epsilon |\mathbf{h}|$$ This means that we can chose $\mathbf{v}=\frac{\mathbf{h}}{|\mathbf{h}|}$ so that $|\mathbf{v}|=1$ and $|A\mathbf{v}|\leq \epsilon$. So some sequence of these $\mathbf{v}$ will comverge say to $\mathbf{w}$ and we will have $|\mathbf{w}|=1$ and $A\mathbf{w}=0$, so $A$ is not injective.


As $a$ is a limit point of $f^{-1}(b)$, we have that for all $r>0$, $B_r(a)\setminus\{a\}\cap f^{-1}(b)\neq\varnothing$. Then exists a injective sequence $(x_n)$ in $f^{-1}(b)$ s.t. $x_n\to a$.

We know from the differentibillity of $f$ that it is continuous, then $b=\lim\limits_{n\to\infty}f(x_n)=f(a)$

As $f$ is differentiable, one can wright $$f(x)=f(a)-f'(a)(x-a)+r(x)=b-f'(a)(x-a)+r(x);$$ with $\lim\limits_{x\to a} \frac{r(x)}{||x-a||}=0$ Then, $$0=f(a)-(x)=f'(a)(x-a)-r(x);$$ that is, $$f'(a)(x_n-a)=r(x_n).$$ Then we know that $$\lim\limits_{n\to\infty}f'(a)\left(\frac{x_n-a}{||x_n-a||}\right)=0.$$ But $$\left\lVert\frac{x_n-a}{||x_n-a||}\right\lVert=1$$ than we have a convergent subsequence $$\frac{x_n-a}{||x_n-a||}\to x;$$ with $||x||=1$. As $f'(a)$ is linear and, therefore, continuous, $$0=\lim\limits_{n\to\infty}f'(a)\left(\frac{x_n-a}{||x_n-a||}\right)=f'(a)x$$ Then $\ker f'(a)\neq\{0\}$, i.e., $f'(a)$ is not injective.


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