TITLE: $\int |f+g|^p \leq \int |f|^p + \int |g|^p$ for $p>0$ and $f, g \in L^p$?
QUESTION [6 upvotes]: $(\Omega, \mathcal{F}, u)$ is a measure space, and $L^p(\Omega, \mathcal{F}, u)$ is its $L^p$ space. Define $N_p(f) = \int_{\Omega} |f|^p\, d\mu$. $\forall f, g \in L^p(\Omega, \mathcal{F}, u)$, 

when $0<p<1$, $N_p(f+g) \leq
    N_p(f)+N_p(g) $ is true according to Wikipedia. This can help to show that     $L^p(\Omega, \mathcal{F}, u)$ is a
vector space. I was wondering how to
prove the inequality is true?
when $p \geq 1$, is the inequality
$N_p(f+g) \leq N_p(f)+N_p(g) $ still true? 
If not, 
(1) can the inequality be modified to be true?
Note I am not asking about the
triangle inequality of $L^p$ norm.
(2) how can one show that $L^p(\Omega, \mathcal{F}, u)$ is a
vector space?

Thanks and regards!

REPLY [6 votes]: This holds because $(a+b)^p\leq a^p+b^p$ when $0<p<1$, $a\geq 0$, and $b\geq 0$, as seen for example here.
Not if $p>1$. For $p\geq 1$ you have $(a+b)^p\leq 2^{p-1}(a^p+b^p)$ when $a\geq0$ and $b\geq 0$.  This can be proved by rearranging the inequality 
$\displaystyle \left(\frac{a+b}{2}\right)^p\leq\frac{1}{2}(a^p+b^p)$, 
which follows from convexity of the function $x\mapsto x^p$.  You need the factor of $2^{p-1}$, as can be seen by taking $a=b=1$.  A less sharp version was asked about here.