Show that the image of a cube is almost a cube

Question

Let $C_r = \left \{ x \in \mathbb{R}^n : |x^i| < r \forall 1 \leq i \leq n \right \}$

and $ g \in C^1(U, \mathbb{R}^n)$ for some open $\left \{ 0 \right \} \subset U$ s.t $dg(\vec{0}) = I, g(\vec{0})=\vec{0}$. and let us choose some $0 < \varepsilon < 1$.

Show that there exists $\delta > 0$ s.t $\forall r < \delta, C_{(1-\varepsilon)r} \subset g(C_{r}) \subset C_{(1+\varepsilon)r}$

I couldn't think of a better title, edits are welcome.

This is what I tried:

$g(0+\triangle x) - g(0) = dg(0)(\triangle x) +o(\triangle x) \implies g(x) = x+o(x) \implies \frac{||g(x) - x||}{||x||} \xrightarrow[x \to 0]{} 0$

So we can choose some $\delta > 0$ s.t if $||x|| < \delta$ then $||g(x)-x|| < ||x||_{\infty} \varepsilon$

Now $x \in C_r \implies |x^i|<r \implies |g(x)^i| < r+ ||x||_{\infty}\varepsilon \leq r+r \varepsilon \implies g(C_{r}) \subset C_{(1+\varepsilon)r}$

But i'm not sure how to show $C_{(1-\varepsilon)r} \subset g(C_{r})$.

Hints appreciated.

Also, does what I did so far seem correct?

Answer 1

Step 1: Definition of multivariate derivative

By definition, $dg(x)$ is the unique matrix which satisfies: $$ \lim_{h\rightarrow 0}\frac{||g(x+h)-(g(x)+dg(x) h)||_2}{||h||_2}=0 $$ In our case, this simplifies to: $$ \lim_{h\rightarrow 0}\frac{||g(h)-h||_2}{||h||_2}=0 $$

Step 2: Working with cubes

But since we are working with cubes, not Euclidean balls, we would perfer the "cube-norm" (aka the sup-norm): $||x||_\infty:=\max_{i\in[n]}|x_i|$. It still works, using the following inequalities (easy to show): $$ ||x||_\infty\leq ||x||_2\leq \sqrt{n}||x||_\infty $$ We can write: $$ 0=\lim_{h\rightarrow 0}\frac{||g(h)-h||_2}{||h||_2}\geq \lim_{h\rightarrow 0}\frac{||g(h)-h||_\infty}{\sqrt{n}||h||_\infty}\geq 0 $$ So, it follows that the limit is zero, also for the $||\cdot||_\infty$-norm. Finally, this means that for any $\varepsilon>0$, there exists a $\delta>0$, such that for all $||h||_\infty<\delta$ (in other words, for all $h\in C_\delta$), $$ \frac{||g(h)-h||_\infty}{||h||_\infty}<\epsilon $$

Step 3: Triangular inequality

Using the triangular inequality, $||h||_\infty\leq ||g(h)-h||_\infty+||g(h)||_\infty$. Therefore: $$ \frac{||h||_\infty}{||h||_\infty}-\frac{||g(h)||_\infty}{||h||_\infty}<\epsilon $$ Rearranging: $$ ||h||_\infty(1-\epsilon)< ||g(h)||_\infty $$ Using instead the triangular inequality $||g(h)||_\infty\leq ||g(h)-h||_\infty+||h||_\infty$, we get: $$ \frac{||g(h)||_\infty}{||h||_\infty}-\frac{||h||_\infty}{||h||_\infty}<\epsilon $$ Which means: $$ ||g(h)||_\infty\leq ||h||_\infty(1+\epsilon) $$ To summarize: $$ ||h||_\infty(1-\epsilon)\leq ||g(h)||_\infty\leq ||h||_\infty(1+\epsilon) $$ This already implies that $g(C_\delta)\subset C_{\delta(1+\epsilon)}$. However, it is not enough to conclude that $g(C_\delta)$ contains all of $g(C_{\delta(1-\epsilon)})$.

Step 3: Inverse function theorem

By the inverse function theorem, there is some open set $V$ containing $0$, such that $g^{-1}$ exists on $V$ and is continuous. Henceforth we will assume $C_{\delta(1+\epsilon)}\subset V$ (if not, we make $\delta>0$ smaller until this is true). Moreover, the inverse function theorem tells us that $dg^{-1}(0)=I^{-1}=I$. Repeat steps 1-2, to obtain: $$ ||g^{-1}(y)-y||_\infty\leq \epsilon||y||_\infty $$ This holds for all $||y||_\infty<\gamma$ for some small $\gamma$, we may take $\gamma=\delta$ without loss of generality.

Step 4: Completing the proof

Take any $x\in C_{\delta(1-\epsilon)}$. We show that $x\in g(C_{\delta})$. Since $g$ is invertible on $C_{\delta}$, this is the same as requiring that $y\in C_{\delta}$ whenever $g^{-1}(y)\in C_{\delta(1-\epsilon)}$. Finally: $$ ||y||_\infty \leq ||g^{-1}(y)||_\infty+||g^{-1}(y)-y||_\infty\leq \delta(1-\epsilon)+\epsilon||y||_\infty $$ Rearranging, we have $||y||_\infty<\delta$. The proof is complete.