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

domath
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    If $x = y \neq 0$ it seems that you have derived $2|x|^p \le 0$ – Calvin Khor Nov 06 '19 at 03:34
  • You need to somehow use the fact that $x=0\implies |x|^p = 0$, I think the result isn't true for general concave functions, see https://math.stackexchange.com/questions/295551/concave-implies-subadditive – Calvin Khor Nov 06 '19 at 03:47

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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 $$ |\alpha (a+b) + (1-\alpha) 0 |^p + |(1-\alpha) (a+b) + \alpha 0 |^p \ge \alpha|a+b|^p + (1-\alpha)|a+b|^p = |a+b|^p. $$

Calvin Khor
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  • Do you remember how to prove the equivalent for $p=1$? Its 100% exactly the same – Calvin Khor Nov 08 '19 at 01:10
  • then if I take $a = f-g$ and $b=g$ It would be $|f-g+g|^p \le |f-g|^p + |g|^p \implies |f|^p - |g|^p \le |f-g|^p $ , what is the role of $0<p<1$ is this proof? – domath Nov 08 '19 at 01:15
  • @stat_yale In the earlier part. For concavity. https://math.stackexchange.com/questions/2678652/can-i-expect-xs-ys-leq-cx-ys-for-s1?noredirect=1&lq=1 see here, result is false for $p>1$ – Calvin Khor Nov 08 '19 at 01:16
  • In above I can write using that 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\ \end{align}

    which is the opposite. What is wrong here?

     – domath Nov 08 '19 at 01:20
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    @stat_yale take $y=-x$ – Calvin Khor Nov 08 '19 at 01:20
  • @stat_yale maybe I should say more. $|t|^p$ is concave $\color{red}{\text{ for }t>0}$. – Calvin Khor Nov 08 '19 at 01:23
  • Right, I got the counter example. Just wondering which step in above is not correct. – domath Nov 08 '19 at 01:23
  • the very first step, application of concavity failed – Calvin Khor Nov 08 '19 at 01:25