Let $(E,\mathcal E)$ be a measurable space, $\kappa$ be a Markov kernel on $(E,\mathcal E)$, $\mu$ be a probability measure on $(E,\mathcal E)$ invariant with respect to $\kappa$, $p\in[1,\infty]$ and $L^p_0(\mu):=\left\{f\in L^p(\mu):\mu f=0\right\}$. As usual, $\kappa f:=\int\kappa(\;\cdot\;,{\rm d}y)f(y)$ and $\mu:=\int f\:{\rm d}\mu$ for $f\in L^1(\mu)$.

How can we show that $\left\|\kappa-\mu\right\|_{\mathfrak L(L^p(\mu))}\le 2\left\|\kappa\right\|_{\mathfrak L(L^p_0(\mu))}$?

The trick seems to be to write \begin{equation}\begin{split}\left\|\kappa-\mu\right\|_{\mathfrak L(L^p(\mu))}&=\sup_{\left\|f\right\|_{L^p(\mu)}\le1}\left\|(\kappa-\mu)f\right\|_{L^p(\mu)}\\&=2\sup_{\left\|f\right\|_{L^p(\mu)}\le1}\left\|\kappa\left(\frac12(f-\mu)\right)\right\|_{L^p(\mu)}\le2\sup_{\substack{\left\|f\right\|_{L^p(\mu)}\le1\\\mu f=0}}\left\|\kappa f\right\|_{L^p(\mu)},\end{split}\tag1\end{equation} but I've no idea why the last inequality holds. Actually, I don't even get the idea behind introducing the factor $2\cdot\frac12$ in the second equality.


1 Answer 1


Simply notice that if $\|f\|_{L^p(\mu)} \leq 1$ then $\|f - \mu(f)\|_{L^p(\mu)} \leq 2$ and also that $\mu(f - \mu(f)) = 0$ so that $\frac12 (f - \mu(f))$ is an element of the set over which you take the $\sup$ on the right hand side of the last inequality.

  • $\begingroup$ Ah, I see. I was thinking in a different direction. Thank you! Do you've got an idea for my related question as well? math.stackexchange.com/q/3432307/47771. $\endgroup$
    – 0xbadf00d
    Nov 12, 2019 at 15:38
  • $\begingroup$ In the special case $p=2$, we should even get $\left\|f\right\|_{L^2(\mu)}^2=\left\|f-\mu f\right\|_{L^2(\mu)}^2+|\mu f|^2$ (noting that $f=(f-\mu f)+\mu f$ is an $L^2(\mu)$-orthogonal sum) and hence $\left\|f-\mu f\right\|_{L^2(\mu)}\le\left\|f\right\|_{L^2(\mu)}$, right? $\endgroup$
    – 0xbadf00d
    Nov 12, 2019 at 16:00
  • 1
    $\begingroup$ @0xbadf00d That's right, yes $\endgroup$ Nov 12, 2019 at 16:05
  • $\begingroup$ I think the claim in the other question is wrong, but this should be correct: math.stackexchange.com/q/3432832/47771. But I'm unsure whether the desired result holds for real Hilbert spaces. $\endgroup$
    – 0xbadf00d
    Nov 12, 2019 at 17:27

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