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.