# Operator norm of symmetric Markov kernel acting on $L^2$

Let $$(S,\mathcal B)$$ be a measurable space, $$\pi$$ be a Markov kernel on $$(S,\mathcal B)$$ and $$\mu$$ be a probability measure on $$(S,\mathcal B)$$ invariant with respect to $$\pi$$. Let $$L^2_0(\mu):=\left\{f\in L^2(\mu):\int f\:{\rm d}\mu=0\right\}=\left\{f\in L^2(\mu):f\perp 1\right\}$$ and $$\gamma(\pi)$$ denote the operator norm of $$\pi f:=\int\pi(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for }f\in L^2_0(\mu),$$ i.e. $$\gamma(\pi)=\left\|\pi\right\|_{\mathfrak L(L^2_0(\mu))}=\sup_{f\perp1}\frac{\left\|\pi f\right\|_{L^2(\mu)}}{\left\|f\right\|_{L^2(\mu)}}.$$

On page 48 of this lecture notes, we can find the following: The last identity seems to be wrong to me. As shown here, we should have $$\gamma(\pi)=\sup_{f\perp1}\frac{\color{red}{|}\langle\pi f,f\rangle_{L^2(\mu)}\color{red}{|}}{\left\|f\right\|_{L^2(\mu)}}\tag1$$ instead. On the other hand, the quantity without the absolute value should be $$\sup_{f\perp1}\frac{\langle\pi f,f\rangle_{L^2(\mu)}}{\left\|f\right\|_{L^2(\mu)}}=\sup_{\lambda\in\sigma(\pi)}\lambda,\tag1$$ where $$\sigma(\pi)$$ denotes the spectrum of $$\pi$$. Am I missing something or is there a mistake in the lecture notes?