The L¹ norm is dominated by the L² norm on a probability space

Cauchy-Schwarz against the constant $1$. On a probability space,

Where this is used. In the truncation argument for the Markov chain central limit theorem, the available control on $Y_n - W^K_n$ is a second-moment bound coming from the $O(n)$ variance estimate for partial sums, whereas the approximation lemma that closes the argument consumes a first-moment bound. This inequality is the bridge, and it is the reason the final approximation error is $2\sqrt{N}\,\|f - f_K\|_{L^2(\pi)}$ rather than something involving an $L^1$ modulus.

Proof. Rather than invoking the Cauchy-Schwarz inequality, use the elementary weighted arithmetic-geometric bound: for every $\varepsilon > 0$ and every real $z$,

which is just $(\varepsilon - |z|)^2 \ge 0$ divided by $2\varepsilon$. Integrating gives $\mathbb E|Z| \le \varepsilon/2 + \mathbb E[Z^2]/(2\varepsilon)$ for every $\varepsilon > 0$. If $S := \mathbb E[Z^2] > 0$, choosing $\varepsilon = \sqrt S$ makes the right-hand side exactly $\sqrt S$. If $S = 0$ then $Z^2 = 0$ almost everywhere, hence $|Z| = 0$ almost everywhere and both sides vanish. (Integrability of $|Z|$ itself follows from that of $Z^2$ on a probability space.)