An upper bound involving the second derivative of the error function

Question

I am trying to bound a function of the form \begin{align} f(x,y) &= \operatorname{erf}(x+y) - 2\operatorname{erf}(x) + \operatorname{erf}(x-y), \end{align} for small values of $y$ and all (sufficiently large) $x$. Obviously this form is rather reminiscent of the symmetric way of writing the second derivative \begin{align} g''(x) = \lim_{y\to 0} \frac{g(x+y) - 2g(x) + g(x-y)}{y^2}, \end{align} so we are motivated to seek a bound that looks like \begin{align} f(x,y) \leq y^2 \frac{d^2}{dx^2} \operatorname{erf}(x) = -\frac{4}{\sqrt{\pi}} xy^2 \exp(-x^2). \end{align} It appears (just numerically plotting it) that this bound is indeed true, as long as $x \geq y$ and $x\geq\sqrt{2}$, which is relevant as the point where the Gaussian $\exp(-x^2)$ switches from being concave to convex.

Unfortunately I have not been able to prove such a bound. I think that such a basic thing should be known somewhere, so hopefully someone here has seen it before!

Answer 1

Denote $G(x) = \text{erf}(x)$. By Taylor's approximation, we have: $$G(x+y) = G(x) +yG'(x)+\frac{y^2}{2}G''(x)+\frac{y^3}{3!}G^{(3)}(x)+\frac{y^4}{4!}G^{(4)}(x) + \mathcal{o}(y^4)$$ $$G(x-y) = G(x) -yG'(x)+\frac{y^2}{2}G''(x)-\frac{y^3}{3!}G^{(3)}(x)+\frac{y^4}{4!}G^{(4)}(x) + \mathcal{o}(y^4)$$ Then we can easily deduce $$\begin{align} f(x,y)&=y^2G''(x)+2\frac{y^4}{4!}G^{(4)}(x) +\mathcal{o}(y^4)\\ &=-\frac{4xy^2}{\sqrt{\pi}}e^{-x^2} \underbrace{-\frac{y^4}{12}\cdot \underbrace{\frac{8x(2x^2-3)}{\sqrt{\pi}}e^{-x^2}}_{\ge0 \text{ for $x\ge \sqrt{3/2}$}}+\underbrace{\mathcal{o}(y^4)}_{\text{dominated by }y^4}}_{\text{negative for $x\ge \sqrt{3/2}$ and $y$ sufficiently small}} \end{align}$$

By consequence, $$f(x,y) \le -\frac{4xy^2}{\sqrt{\pi}}e^{-x^2} \hspace{1cm} \text{for all $x\ge \sqrt{3/2}$ and $y$ sufficiently small}$$

Answer 2

Some thoughts.

Remarks: We can prove that if $x \ge \frac32$ and $x \ge y > 0$, the desired inequality is true. I think that $x \ge \frac32$ can be relaxed to $x \ge \sqrt{2}$.

Fact 1: Let $x \ge \frac32$ and $x \ge y > 0$. Then $$ -\frac{4}{\sqrt{\pi}} xy^2 \exp(-x^2) \ge \operatorname{erf}(x+y) - 2\operatorname{erf}(x) + \operatorname{erf}(x-y).$$

Proof of Fact 1.

Let $$F(y) := -\frac{4}{\sqrt{\pi}} xy^2 \exp(-x^2) - \operatorname{erf}(x+y) + 2\operatorname{erf}(x) - \operatorname{erf}(x-y).$$ We have, for all $y \in (0, x]$, $$F'(y) = \frac{4xy}{\sqrt{\pi}}\exp(-x^2 - y^2) \Big(\frac{\exp(2xy) - \exp(-2xy)}{2xy} - 2\exp(y^2)\Big) \ge 0. \tag{1}$$ (The proof is given at the end.)

Also, we have $F(0) = 0$. Thus, $F(y) \ge 0$ for all $y \in (0, x]$ with $x \ge 3/2$.

We are done.

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Proof of (1).

It suffices to prove that $$\frac{\exp(2xy) - \exp(-2xy)}{2xy} - 2\exp(y^2) \ge 0. \tag{A1}$$

Let $g(u) := \frac{\mathrm{e}^u - \mathrm{e}^{-u}}{u}$. We have $$g'(u) = \frac{\mathrm{e}^{-u}}{u^2} [1 + u - (1 - u)\mathrm{e}^{2u}] \ge 0. \tag{A2}$$ (Note: We only need to prove the case that $0 < u < 1$. It suffices to prove that $H(u) := \ln\frac{1 + u}{1 - u} - 2u \ge 0$. We have $H'(u) = \frac{2u^2}{1 - u^2} \ge 0$. Also, $H(0) = 0$. Thus, $H(u) \ge 0$ for all $0 < u < 1$.)

If $y \ge 3/2$, from (A1) and (A2), using $x \ge y$, it suffices to prove that $$\frac{\exp(2\cdot y \cdot y) - \exp(-2\cdot y \cdot y)}{2\cdot y \cdot y} - 2\exp(y^2) \ge 0 \tag{A3}$$ which is true.
(Note: Using $1 + (y^2 - 9/4) \le \mathrm{e}^{y^2 - 9/4} \le \frac{3}{28}\mathrm{e}^{y^2}$. It suffices to prove that $\frac{\exp(2\cdot y \cdot y) - \exp(-2\cdot y \cdot y)}{2\cdot (\frac{3}{28}\exp(y^2) + \frac54) } - 2\exp(y^2) \ge 0$. Letting $v = \exp(y^2) \ge \exp(9/4)$, it suffices to prove that $\frac{v^2 - 1/v^2}{2(\frac{3}{28}v + \frac54)} - 2v \ge 0$ which is true. )

If $0 < y < 3/2$, from (A1) and (A2), using $x \ge 3/2$, it suffices to prove that $$\frac{\exp(2\cdot \frac32 \cdot y) - \exp(-2\cdot \frac32 \cdot y)}{2\cdot \frac32 \cdot y} - 2\exp(y^2) \ge 0 \tag{A4}$$ which is true.
(Note: If $0 < y < 1/3$, use $\exp(y^2) \le 1 + \frac32 y^2$. If $1/3 \le y < 3/2$, use $\exp(y^2) \le \exp(3y/2)\cdot (3y^2/4 - 9y/8 + 1)$.)

We are done.