One version of mean value functional (nonadditive Markov operator) $\mathcal{M} \colon C(\Omega)_+ \to C(\Omega)_+ $ is defined as follows $$( \mathcal{M} f ) (x) = \left\lbrace \int_{\Omega} f(y)^{p} \pi(\mathrm{d} y | x) \right\rbrace^{1/p} $$ where $\pi(\mathrm{d} y | x)$ is a weakly continuous stochastic (Markov) kernel.

Alternatively, we define $\mathcal{M} \colon C(\Omega)_+ \to C(\Omega)_+$ as $$( \mathcal{M} f ) (x) = \left\lbrace \int_{\Omega} f(y)^{p} k(x, y) \mathrm{d} y \right\rbrace^{1/p} $$ where $k (x, y) \colon \Omega \times \Omega \to \mathbb{R}_+$ is any weakly continuous stochastic density kernel.

Suppose that $f$ and $g$ are two continuous real-valued nonnegative functions defined on a Polish space $\Omega$.

I am wondering that is it still correct to use the well-known result of Minkowski's inequality to show $$ \mathcal{M} (f + g ) (x) \leq \mathcal{M} f (x) + \mathcal{M}g (x), \qquad \forall x \in \Omega$$ holds whenever $ p \geq 1$, while the reversed inequality holds whenever $p\leq 1$?

I'm also curious that if above answer is no, then does the answer depend on the probability measure and how?

Any suggestions are much appreciated! Thanks in advance!

  • $\begingroup$ If you put $d\mu= k(x,y)dy$. Then it should be Ok. $\endgroup$
    – Ice sea
    Sep 23 '17 at 6:42

It's worth looking at the general form of Minkowski's inequality: if $\mu$ is any measure on $\Omega$, then $$ \left(\int_\Omega|f(y)+g(y)|^p\mu(dy)\right)^{1/p}\le\left(\int_\Omega|f(y)|^p\mu(dy)\right)^{1/p}+\left(\int_\Omega|g(y)|^p\mu(dy)\right)^{1/p}.$$ In your case, simply use $\mu(dy)=\pi(dy|x)$ (or $\mu(dy)=k(x,y)dy$ in the second case, but this is just a special case of the first). Note that$f,g\ge0$ so you don't need to worry about the absolute value.

  • $\begingroup$ Thanks a million Jason:) May I confirm your answer that it is correct to directly apply the general form of Minkowski's inequality to non-additive Markov operator by using $\mu (dy) = \pi (dy |x)$ ? To apply it, is there any additional restrictions (requirement) on the probability measure $\pi (dy|x)$? Cheers:) $\endgroup$ Sep 23 '17 at 23:44
  • 1
    $\begingroup$ No restrictions, the inequality I have written is true for any real (or complex) valued functions $f,g$ on any measure space $(\Omega,\mathcal F,\mu)$. $\endgroup$
    – Jason
    Sep 23 '17 at 23:45

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