show that $| f+g | ^p \leq s^{1-p}| f| ^p + (1-s)^{1-p}| g| ^p$

Question

I would like to show that $$| f+g | ^p \leq s^{1-p}| f| ^p + (1-s)^{1-p}| g| ^p$$ for all $0<s<1$ and $p\geq 1$, with $f$ and $g$ to functions defined on $\mathbb R^n$ with values in $\mathbb R$. I would like to use the convexity of $t \mapsto {| t|}^p$ but I don't know I to do it ...

Answer 1

Convexity says that for any $s\in(0,1)$, $F\in\mathbb R$, and $G\in\mathbb R$, $$ |s F + (1-s)G|^p \le s|F|^p + (1-s)|G|^p.$$ To prove the result you want, simply apply the above result with $$f=s F,\quad g= (1-s)G.$$ Since $$ |F|^p = s^{-p}|f|^p, \quad |G|^p = (1-s)^{-p}|g|^p,$$ this immediately implies the result.

Answer 2

Hints: Let $a, b \geq 0$ and define $f(s)=s^{1-p} a^{p}+(1-s)^{1-p} b^{p} -(a+b)^{p}$. Check that $f'(s)=0$ gives $s =\frac a {a+b}$. Then verify that $f(s)=0$. This gives $f(s) \geq 0$ for all $s \in (0,1)$. Now take $a=|f(x)|, b=|g(x)|$ to get the result. [Note that $|f(x)+g(x)| \leq a+b$].