= Solution
Subtract a constant so that $f(0)=0$, and write $m=\mathbb Ef(Y)$, $V=\operatorname{Var}f(Y)$, and $A=\mathbb E f'(Y)^2$. The supplied identity applied to $f$ and $f^2$ gives
$$
m=\mathbb E[\operatorname{sgn}(Y)f'(Y)],
\qquad
\mathbb Ef(Y)^2=2\mathbb E[\operatorname{sgn}(Y)f(Y)f'(Y)].
$$
By <Cauchy-Schwarz inequality>, $m^2\leq A$ and
$$
V=\mathbb Ef^2-m^2
\leq2\sqrt{(V+m^2)A}-m^2.
$$
Thus $V+m^2\leq2\sqrt{(V+m^2)A}$, so $V+m^2\leq4A$ and in particular $V\leq4A$. Therefore the standard Laplace distribution has $C_P(Y)\leq4$.
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