# Show that the image of a cube is almost a cube

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|

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?

• I now notice I made a mistake as i need $|g(x)^i| < (1+\varepsilon)r$ and I showed something else. Feb 24, 2021 at 17:55
• Remark: Either work with a more suitable norm or use the usual norm and balls (with cubes inscribed and superscribed). Feb 24, 2021 at 19:06
• @TedShifrin Thanks. I fixed the mistake, still looking for help on the other part of the question. Feb 25, 2021 at 23:59
• How did you prove the open mapping part of the Inverse Function Theorem? Feb 26, 2021 at 1:18
• I tried looking at the proof now but I'm not sure how this helps... Feb 26, 2021 at 12:31

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.