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||$

Question

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??

Answer 1

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 $0 < \lVert h \rVert < \delta$?

Answer 2

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)