$\int \bigg||f|^p-|g|^p\bigg|d\mu\leq \int \big|f-g\big|^pd\mu$ , for $ 0<p<1$

Question

Prove that $$\int \bigg||f|^p-|g|^p\bigg|d\mu\leq \int \big|f-g\big|^pd\mu$$

for all $f, g \in L^p([0, 1]; \mu)$ and $0 < p < 1$.

Is the following correct?

the function $f(t)=|t|^p$ , $0 < p < 1$ is concave. So $\forall \alpha \in (0,1)$;

$$ |x-y|^p = |\alpha\frac{x}{\alpha}-(1-\alpha)\frac{y}{1-\alpha}|^p \geq \alpha|\frac{x}{\alpha}|^p+(1-\alpha)|\frac{y}{1-\alpha}|^p$$

let $\alpha = \frac{|x|}{|x|+|y|}$ and $1-\alpha = \frac{|y|}{|x|+|y|}$ \begin{align} |x-y|^p & \geq \alpha\frac{|x|^p}{ \frac{|x|^p}{(|x|+|y|)^p}}+(1-\alpha)\frac{|y|^p}{ \frac{|y|^p}{(|x|+|y|)^p}}\\ & = \alpha(|x|+|y|)^p+(1-\alpha)(|x|+|y|)^p\\ & = (|x|+|y|)^p\\ & \geq |x|^p+|y|^p\\ & \geq \bigg||x|^p-|y|^p\bigg|\\ \end{align}

so taking integration $$|f-g|^p \geq \bigg||f|^p-|g|^p\bigg|$$ $$\int|f-g|^pd\mu \geq \int\bigg||f|^p-|g|^p\bigg|d\mu$$

Answer 1

Its enough to prove the identity for numbers $a,b\ge 0$, $$ |a+b|^p + \le |a|^p + |b|^p$$

We use the idea from Concave implies subadditive.

Note that $a = \alpha (a+b)$, $b = (1-\alpha)(a+b)$, where $\alpha = \frac{a}{a+b}$. Then the right hand side is $$ |\alpha (a+b) + (1-\alpha) 0 |^p + |(1-\alpha) (a+b) + \alpha 0 |^p \ge \alpha|a+b|^p + (1-\alpha)|a+b|^p = |a+b|^p. $$