Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.