$$Cov(f(X), g(Y)) \leq \vert \beta \vert \sqrt{Var(f(X)) Var(g(Y))}$$.
In fact, one can define the maximal correlation of random variable by the best constant above and of course it should be bigger than $\beta$. Let us remark how to prove the inequality above quickly. We can use the expansion by Hermit polynomial that we have
$$\mathbb{E}[H_n(X) H_m(Y)] = \delta_{n,m} \left(\mathbb{E}\frac{1}{n!}[XY]\right)^n.$$
Then a centered $L^2$ functions have projection on $H_0$ zero. Then we have
$$\mathbb{E}[f(X)g(Y)] = \sum_{n=1}^{\infty}\langle f, H_n \rangle \langle g, H_n \rangle \frac{1}{n!} \beta^n \\ \leq \vert \beta \vert \sqrt{\sum_{n=1}^{\infty}\langle f, H_n \rangle^2 \frac{1}{n!} } \sqrt{\sum_{n=1}^{\infty}\langle g, H_n \rangle^2 \frac{1}{n!} } \\ \leq \beta \vert \sqrt{Var(f(X)) Var(g(Y))}.$$
This concludes the proof.
This concludes the proof.
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