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 $x \in \mathbb{R}^n$?

  • $\begingroup$ Is $\Vert\Vert$ the standard Euclidean norm, or any norm or $\mathbb{R}^n?$ $\endgroup$ – MSDG Oct 25 '18 at 21:50
  • $\begingroup$ $\mathbb{R}^n$. 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 $\left\lVert * \right\rVert_\infty$ $\endgroup$ – JavaTeachMe2018 Oct 25 '18 at 21:58
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    $\begingroup$ In that case, this link might be helpful. $\endgroup$ – 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 $\frac x {\|x\|_{\infty}}$.

  • $\begingroup$ 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... $\endgroup$ – JavaTeachMe2018 Oct 26 '18 at 16:05
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Oct 26 '18 at 23:09

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