# Assuming $A = Df(x_0)$ is invertible, prove that there exists $\mu > 0$ such that for all $x \in R^n$ $||Ax|| \geq \mu||x||$

My questions are

Let $$U \subset R^n$$ be an open set, $$f: U \rightarrow R^n$$ be a $$C^1(U)$$ function and $$x_0 \in U$$.

1) Assuming $$A = Df(x_0)$$ is invertible, prove that there exists $$\mu > 0$$ such that for all $$x \in R^n$$ $$||Ax|| \geq \mu||x||$$

2) Prove there exist a $$\delta > 0$$ such that $$f$$ on $$B_{\delta}(x_0)$$ = $${ y \in R^n: ||y-x_o|| < \delta}$$ such that f is one-to-one.

For 1), I think I have some scratch work of " Supposing $$Ax =0$$ then $$||x|| = 0$$ but I don't really understand it clearly.

For 2), it seems like something to do with the inverse function theorem??

• Do you know the operator norm? Could be use this for (1). Also, needs to clarify what do you want in (2). – azifmedrano Mar 14 at 5:51
• I am not familiar with the operator norm, could you explain a little more? I also clarified #2. – pop Mar 14 at 6:06

It is a good exercise for you to prove the following lemma:

If $$T : \Bbb{R}^n \to \Bbb{R}^n$$ is a linear transformation, then, there exists a constant $$\nu$$ such that for all $$x \in \Bbb{R}^n$$, \begin{align} \lVert T(x) \rVert \leq \nu \lVert x\rVert \end{align} In fact, one may take $$\nu = \sup\{\lVert T(x)\rVert: \, \, x \in \Bbb{R}^n, \, \, \lVert x \rVert = 1 \}$$.

Often, this supremum is called the operator norm of $$T$$, and is denoted $$\lVert T\rVert$$. So, $$\lVert T(x) \rVert \leq \lVert T \rVert \cdot \lVert x \rVert.$$ (if you really get stuck, you could probably search on this site, or I could elaborate more, but really try it yourself first).

Having established the lemma, apply it to $$T = A^{-1}$$, and try to justify why the constant $$\mu$$ you get is strictly positive.

For $$(2)$$, indeed, the inverse function theorem gives you this result for free (it states even more than what $$(2)$$ says). However, $$(2)$$ is usually used as a lemma in proving the inverse function theorem, so it is a good idea to prove it independently as well. The key is to make use of (1). Note that by definition of differentiability of $$f$$ at $$x_0$$, we have that \begin{align} \lim_{h \to 0} \dfrac{\lVert f(x_0 + h) - f(x_0) - A(h)\rVert}{\lVert h \rVert} = 0 \end{align}

This means for every $$\epsilon > 0$$, there exists a $$\delta > 0$$, such that the open ball $$B_{\delta}(x_0)$$ is contained in $$U$$ (because it is open) and such that for all $$h \in B_{\delta}(x_0)$$, we have that \begin{align} \lVert f(x_0+ h) - f(x_0) - A(h)\rVert \leq \epsilon\lVert h \rVert. \end{align}

So, now using the $$\mu$$ you find from (1), choose $$\epsilon$$ such that $$0 < \epsilon < \mu$$ (for example $$\epsilon = \mu/2$$); then there is a $$\delta$$ as in the above remark. Now, using the (reverse) triangle inequality, we find that \begin{align} \lVert A(h) \rVert - \lVert f(x_0+h) - f(x_0)\rVert \leq \epsilon \lVert h \rVert \end{align} Using the fact that $$\lVert A(h) \rVert \geq \mu \lVert h\rVert$$, and rearranging the above inequality, we find that \begin{align} \underbrace{(\mu - \epsilon)}_{>0}\lVert h \rVert \leq \lVert f(x_0 + h) - f(x_0) \rVert \end{align} Recall once again that this is true for all $$h$$ such that $$0 \leq \lVert h \rVert < \delta$$. What can you conclude if $$0 < \lVert h \rVert < \delta$$?

• Thank you for a thoughtful and thorough answer! I'd really appreciate it. – pop Mar 14 at 6:36

The second question can be answered using the inverse function theorem. Since you have already stated that f is $$C^1$$, and that $$Df(x_0)$$ is invertible, by the InvFT one can conclude that there exits open neighbourhoods of $$x_0$$ and $$y_0$$, say $$N_{x_0}$$ and $$N_{y_0}$$ such that $$f:N_{x_0}\rightarrow N_{y_0}$$ is bijective (and hence one-to-one). Since $$N_{x_0}$$ is open and $$x_0\in N_{x_0}$$, there exists some $$\delta>0$$ such that $$\mathbb{B}_{\delta}(x_0)\subseteq N_{x_0}$$. So, $$f$$ on $$\mathbb{B}_{\delta}(x_0)$$ is bijective (and hence one-to-one)