# Proving $f: \mathbb{R}^n \ni x\to \left\lVert x \right\rVert$ $\in \mathbb{R}$ to be continuous

I had to prove that $$f: \mathbb{R}^n \ni x\to\left\lVert x \right\rVert$$ $$\in \mathbb{R}$$ is continuous regarding maximum norm.

Is it correct to do the following, i.e. using triangle inequality?

$$\left\lVert x \right\rVert$$ = $$\left\lVert x - y + y\right\rVert$$ $$\leq$$ $$\left\lVert x - y \right\rVert$$ $$+ \left\lVert y \right\rVert$$

$$\left\lVert y \right\rVert$$ = $$\left\lVert y - x + x\right\rVert$$ $$\leq$$ $$\left\lVert y - x \right\rVert$$ $$+ \left\lVert x \right\rVert$$

After rearranging I get

$$|f(x) - f(y) |$$ $$\leq$$ $$\left\lVert x-y \right\rVert$$

which is why f is Lipschitz continuous, and therefore continuous. (I think?)

Now there is $$S := {x \in \mathbb{R}^n}$$ : $$\left\lVert x \right\rVert_\infty = 1$$ which is the unit sphere regarding maximum norm. How can I now conclude from above, that $$f$$ has a minimum $$A := f(x) > 0$$ in a point $$x$$ on $$S$$?

And is it also possible to conclude out of that, that therefore $$\left\lVert x\right\rVert$$ $$\geq$$ $$A * \left\lVert x \right\rVert_\infty$$ for all $$x \in \mathbb{R}^n$$?

• Is $\Vert\Vert$ the standard Euclidean norm, or any norm or $\mathbb{R}^n?$ – MSDG Oct 25 '18 at 21:50
• $\mathbb{R}^n$. Basically the task was to prove step-by-step that all norms are equivalent in $\mathbb{R}^n$, which is possible by proving that every random norm $\left\lVert * \right\rVert$ in $\mathbb{R}^n$ is equivalent to the max norm $\left\lVert * \right\rVert_\infty$ – JavaTeachMe2018 Oct 25 '18 at 21:58
• In that case, this link might be helpful. – MSDG Oct 25 '18 at 22:00

You did not prove continuity of $$f$$ w.r.t. $$\|x\|_{\infty}$$ norm. Write any vector $$x$$ as $$\sum x_ie_i$$ where $$e_1,e_2,..,e_n$$ is the standard basis. Then $$f(x) \leq \sum |x_i| |f(e_i)|\leq \|x\|_{\infty} \sum |f(e_i)|$$ which proves that $$f$$ is continuous for the $$\|x\|_{\infty}$$ norm. Since $$S$$ is compact in $$\|x\|_{\infty}$$ norm it follows that $$f$$ has a minimum value $$c$$ on $$S$$. [Observe that $$f$$ does not vanish at any point of $$S$$]. Now you can verify that $$f(x) \geq c \|x\|_{\infty}$$ for all $$x$$ by considering $$\frac x {\|x\|_{\infty}}$$.
• Hi, thanks for your answer. Can you evaluate on the last point? I don't understand how I can verify that $f(x) \geq c$ by considering what you have written... – JavaTeachMe2018 Oct 26 '18 at 16:05
• @JavaTeachMe2018 Since $f$ is a norm, $f(x)=0$ implies $x=0$. Since it is continuous on the compact set $S$, its minimum value $c$ is positive. – Kavi Rama Murthy Oct 26 '18 at 23:09