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Let $f(x)$ denote any vector norm. Show that $f(x)$ is continuous with respect to the $1$-norm.

I'm using the $\varepsilon$-$\delta$ approach:

$f(x)$ is continuous at $x_0$ if, given any $\varepsilon>0 $ there exists a $\delta >0$ such that if

$$\|x-x_0\|_1 < \delta \Rightarrow |f(x)-f(x_0)|< \varepsilon$$

Is that what is meant by "with respect to the $1$-norm?" If it is, then my second question is how is this evaluated? I have tried with the triangle inequality and the similarity of norms but can't get anywhere. Any help is appreciated, thanks.

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    $\begingroup$ Presumably this is for finite dimensions. $\endgroup$ – copper.hat Sep 5 '17 at 23:32
  • $\begingroup$ Replace the norm on $f$ by absolute value. $\endgroup$ – copper.hat Sep 6 '17 at 0:06
  • $\begingroup$ @copper.hat yes for finite. Just replace it? If so what is the point on mentioning "w.r.t 1-norm? " $\endgroup$ – JustANoob Sep 6 '17 at 0:17
  • $\begingroup$ $f$ has values in $[0,\infty)$, the norm $\|\cdot \|_1$ is meaningless in that context. $\endgroup$ – copper.hat Sep 6 '17 at 0:20
  • $\begingroup$ The one norm applies to the domain here, implicitly the absolute value $|\cdot|$ is used on the range. $\endgroup$ – copper.hat Sep 6 '17 at 1:22
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Big hint: $f(x) \le \sum_k |[x]_k| f(e_k)$.

Another hint: $|f(x)-f(y)| \le f(x-y)$.

Combining: $|f(x)-f(y)| \le f(x-y) \le \sum_k |[x]_k-[y]_k| f(e_k) \le M \sum_k |[x]_k-[y]_k| = M \|x-y\|_1$, where $M= \max_k f(e_k)$.

Choose $\delta = {1 \over M} \epsilon$.

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  • $\begingroup$ What does the $e_k $ stand for? $\endgroup$ – JustANoob Sep 6 '17 at 8:26
  • $\begingroup$ The $k$th unit vector. $\endgroup$ – copper.hat Sep 6 '17 at 16:29
  • $\begingroup$ Dont get anywhere... $\endgroup$ – JustANoob Sep 6 '17 at 20:04

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