Upper bound for $|\mathbb E[f(X)] - f(0)|$ when $X \sim N(0,\sigma^2)$ and $f:\mathbb R \to \mathbb R$ is piecewise-linear with $\|f'\|_\infty \le 1$.

Question

Let $f:\mathbb R \to \mathbb R$ be a piecewise linear function such that $f(0)=0$ and $\|f'\|_\infty \le 1$ and let $X$ be a Guassian random variable with mean $0$ and variance $\sigma^2$. What is an upper-bound for $|\mathbb E[f(X)]|$ ?

Example

If $f(x) = \max(x,0)$, then $\mathbb E[f(X)]=\sigma/\sqrt{2\pi} = \mathcal O(\sigma)$.

Answer 1

There needs to be more constraints for it to have a bound independent of $f$.

Consider: $f(x) = a - x$ for some large constant $a$. Then $\mathbb{E}[f(X)] = O(a)$ (e.g., when $a >> \sigma$).

Edit: With the new constraint, doubling the bound in the example (given in the question) gives a bound.

Answer 2

Turns out one can get a bound for general $f \in \mathcal C^1$ for which $\|f'\|_\infty \le 1$, via the Mean-Value Theorem and the Cauchy-Schwarz inequality for expectations.

---

Indeed, the MVT gives $f(X) = \pm Xf'(\theta(X))$ for some $\theta(X)$ between $\min(X,0)$ and $\max(X,0)$. Here $\theta(X)$ is a random variable which depends measurably on $X$. Thus $$ |\mathbb E[f(X)]| \overset{\text{MVT}}{=} |\mathbb E[Xf'(\theta(X))]| \overset{\text{C.S}}{\le} \sqrt{\mathbb E[X^2]}\sqrt{\mathbb E[(f'(\theta(X)))^2]} \le \sqrt{\mathbb E[X^2]} = \sigma. $$