# How can we show this inequality for the operator norm of a Markov kernel acting on $L^p$?

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

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.
• 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? Nov 12, 2019 at 16:00