TITLE: $l^p$ not norm, $p<1$
QUESTION [5 upvotes]: please I tried to find counterexamples to see that $l^p$ is not norm with $p<1$ in the triangle inequality but I have problems with convergence when I choose some successions. Thanks.

REPLY [5 votes]: Let $p<1$. Then 
$$\|(1,0,0,\ldots,0\ldots)\|_p=1$$
$$\|(0,1,0,\ldots,0\ldots)\|_p=1$$
$$\|(1,1,0,\ldots,0\ldots)\|_p=2^{1/p}>2$$

REPLY [3 votes]: The triangle inequality is violated for $v_1=(1,0,0,...)$ $v_2=(0,1,0,0,...)$.