Show that any norm is continuous w.r.t the 1-norm

Question

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.

Answer 1

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