Lemma regarding variance and Lipschitz functions

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I'd like to be sure that this lemma is true:

Let $X$ be a real valued random variable defined on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Let $F:\mathbb{R} \rightarrow \mathbb{R}$ be a Lipschitz function of constant $M$:

\begin{equation*} \lvert F(x)-F(y) \rvert \le M \lvert x-y \rvert \quad \forall x,y \in \mathbb{R} \end{equation*} Then \begin{equation*} \lvert F(\mathbb{E}(X))-\mathbb{E}(F(X))\rvert \le M\sqrt{\mathbb{V}ar(X)} \end{equation*}

A proof:

\begin{eqnarray*} \lvert F(\mathbb{E}(X))-\mathbb{E}(F(X))\rvert &\le& \lvert \mathbb{E}(F(\mathbb{E}(X))-F(X))\rvert\\ &\le& \mathbb{E}(\lvert F(\mathbb{E}(X))-F(X) \rvert)\\ &\le& M\mathbb{E}(\lvert \mathbb{E}(X)-X\rvert) \end{eqnarray*}
and an application of Cauchy-Schwarz inequality proves the result.

Is that correct?

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What you have so far is correct. To conclude, you can use Cauchy-Schwarz as you suggested, or monotonicity of $L^p$ norms as follows: $$E[|E[X]-X|] = || X-E[X] ||_{L^1} \leq || X- E[X] ||_{L^2} = E[|X-E[X]|^2]^{1/2} = \sqrt{\mathrm{Var}(X)}.$$