Real Analysis, Folland Problem 6.1.16 $L^p$ spaces

Question

Problem 6.1.16 - If $0 < p < 1$, the formula $\rho(f,g) = \int |f-g|^p$ defines a metric on $L^p$ that makes $L^p$ into a complete topological vector space.

Attempted proof - Suppose $a,b > 0$ and $0 < p < 1$. For $t > 0$ we have $t^{p-1} > (a + t)^{p-1}$, and by integrating from $0$ to $b$ we obtain $a^p + b^p > (a+b)^p$. Let $E$ and $F$ be disjoint sets of positive finite measure in $X$ and set $a = \mu(E)^{1/p}$ and $b = \mu(F)^{1/p}$ we see that $$\|1_{E} + 1_{F}\|_{p} = (a^p + b^p) > a + b = \|1_{E}\|_{p} + \|1_{F}\|_{p}$$

I am not sure where to go from here, any suggestions is greatly appreciated.

Answer 1

Claim: If $0 < p < 1$ then $(x + y)^p \le x^p + y^p$ for $x,y \ge 0$.

The proof is the standard one for this type of inequalities: following PhoemueX's hint, define $$f(y) = (1 + y)^p - 1 - y^p.$$ Notice that $f(0) = 0$, $f'(y) \le 0$ for $y > 0$ and conclude that $f(y) \le 0$. This implies the claim and hence that $\rho(f,g)$ is a metric (the other two properties are trivial to check).

Claim: $(L^p,\rho)$ is complete.

Follow step by step the proof of completeness for $p \ge 1$, the argument is almost exactly the same.