1-norm less than or equal to n times infinity norm

Question

$$||v||_1 \le n||v||_\infty$$

I've been unable to prove this or find a relevant proof. My inexperience with summations (n is the summation limit from the norm definition) is likely to blame. Any help would be much appreciated.

$$\sum_{i=1}^n |v_i| \le n(\sum_{i=1}^n |v_i|^\infty)^{1/\infty} $$

Answer 1

The infinity norm is nothing more than the largest absolute value present among the entries of your vector. Clearly the sum of $n$ copies of the largest value (in other words, $n$ times the largest value) is greater than or equal to the sum of the $n$ values.

Answer 2

To write what Arthur did a little more verbosely, we have that: $$||v||_{\infty} = \max_i|v_i|$$ where $v = (v_1,v_2,\dots,v_n)$. So, we have that: $$n||v||_\infty = n\max_i|v_i| = \sum_{i = 1}^n \max_i|v_i|$$ Now, we clearly have that: $$\sum_{i = 1}^n |v_i|\leq\sum_{i = 1}^n \max_i|v_i|$$ so, we have that: $$||v||_1\leq n||v||_\infty$$