# Finding the maximum of $|\widehat{f''}|$ for $f$ in terms of the Gaussian

Let $$f = \begin{cases} e^{-x^2/2} - e^{-2 x^2} &\text{if x\geq 0,}\\ 0 &\text{if x<0.}\end{cases}$$ I would like to find out $|\widehat{f''}|_\infty$. A good numerical bound -- of preference, an estimate with a very small, explicit error term -- would do, provided that it is rigorous.

It is not that I can't figure out how to do this, but rather that, since this is not really my sort of thing, I am probably doing it in a way that is much too roundabout and otherwise non-exemplary. This is an issue, since this will go into a text that will be read by graduate students (among others) and I do not wish to corrupt the young.

I'll leave my own method below, as a self-answer. Can you do better? How?

Let $$g_a = \begin{cases} e^{-a x^2} & \text{if x\geq 0},\\ 0 &\text{if x<0.}\end{cases}$$ Using the definition of the Fourier transform, completing the square and shifting the contour of integration, $$\widehat{g_a}(t) = \int_0^\infty e^{- a x^2} e^{-2 \pi i x t} dx = e^{-\pi^2 t^2/a} \left(-\int_0^{\frac{i \pi t}{a}} e^{- a z^2} dz + \int_0^\infty e^{- a z^2} dz\right).$$ The last integral equals $\sqrt{\pi/a}/2$. We can thus write $$\widehat{g_a}(t) = \frac{\sqrt{\pi}}{2\sqrt{a}} e^{-\frac{\pi^2}{a} t^2} - \frac{i}{\sqrt{a}} F\left(\frac{\pi t}{\sqrt{a}}\right) = \frac{\sqrt{\pi}}{2\sqrt{a}} e^{-\frac{\pi^2}{a} t^2} \left(1 - \text{erf}\left(\frac{i \pi t}{\sqrt{a}}\right)\right),$$ where $F(x) = e^{-x^2} \int_0^x e^{y^2} dy$ (Dawson's integral) and $\text{erf}(s) = (2/\sqrt{\pi}) \int_0^s e^{-s^2} dz$ (the error function).

Of course, $$\widehat{f''}(t) = (2\pi i t)^2 \widehat{f}(t) = - 2 \pi^{5/2} t^2 (\widehat{g_{1/2}}(t) - \widehat{g_2}(t)).$$

Before we proceed, note that we can easily give a somewhat crude bound that is useful for $t$ large: since $f''$ has a jump (from $0$ to $3$) at the origin, but is differentiable elsewhere, $$|\widehat{f''}(t)|\leq \frac{|\widehat{f'''}(t)|}{2\pi |t|} \leq \frac{|\widehat{f'''}|_1}{2\pi |t|} = \frac{1}{2\pi |t|} \left(3 + \lim_{x_0\to 0^+} \int_{x_0}^\infty |f'''(x)| dx\right),$$ where $f'''$ is to be understood as a distribution (more precisely: a point mass at the origin plus a function to the right of $0$). Since we are deriving a crude bound for now, we can use the inequality $|f'''(x)|\leq |g_{1/2}'''(x)| + |g_{2}'''(x)|$: $$\lim_{x_0\to 0^+} \int_{x_0}^\infty |f'''(x)| dx = \lim_{x_0\to 0^+} \int_{x_0}^\infty |g_{1/2}'''(x)| dx + \lim_{x_0\to 0^+} \int_{x_0}^\infty |g_{2}'''(x)| dx.$$ Now, we can easily see that $g_a'''(x) = (- 8 a^3 x^3 + 12 a^2 x) e^{- a x^2}$ is positive for $0<x<\sqrt{3/2a}$ and negative for $x>\sqrt{3/2a}$, and that $g_a''(0)= -2a$ and $\lim_{x\to \infty} g_a''(x) = 0$. Hence $$\lim_{x_0\to 0^+} \int_{x_0}^\infty |g_{a}'''(x)| dx = 2 a + 2 |g_a''(\sqrt{3/2 a})| = 2 a + 4 a e^{-3/2},$$ and so $$\lim_{x_0\to 0^+} \int_{x_0}^\infty |f'''(x)| dx = 2 (1/2 + 2) + 4 (1/2 + 2) e^{-3/2} = 5 + 10 e^{-3/2}.$$ We conclude that $$|\widehat{f''}(t)|\leq \frac{4 + 5 e^{-3/2}}{\pi |t|}.$$ We will use this for $t>2$, say. Now we need to find the maximum of $|\widehat{f''}(t)|$ for $0<t\leq 2$, and to show that it is greater than $(4+5 e^{-3/2})/2\pi = 0.81418\dotsc$.

In the future, we will no doubt have standard libraries that give values for $\text{erf}$ (for imaginary argument) or $F$, correctly rounded, or within interval arithmetic. AFAIK, that is unfortunately not quite the case yet. (See Chevillard and Revol, "Computation of the error function erf..." for a discussion of erf with real argument.) Thus, we have to do things ourselves, using the Taylor expansion $$\text{erf}(ix) = \frac{2}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{(-1)^n (i x)^{2n+1}}{n! (2n+1)} = \frac{2 i}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{x^{2n+1}}{n! (2n+1)} = \frac{2 i}{\sqrt{\pi}} \left(\sum_{n=0}^{N-1} \frac{x^{2n+1}}{n! (2n+1)} + O^*\left(\frac{x^{2N+1}/(N! (2N+1))}{1- \frac{x^2}{N+1}}\right)\right),$$ where $O^*(\epsilon)$ denotes a quantity with absolute value at most $\epsilon$.

For $x\leq \pi t \cdot \sqrt{2}\leq 3 \sqrt{2} \pi$, we can use this series with a depressingly large $N$ ($N=1000$, say) to get a reasonable approximation to $\text{erf}(i x)$ (using interval arithmetic as an easy way to keep matters rigorous). This gives us an approximation to $\widehat{f''}(t)$ for $t\leq 2$. We can then use the bisection method (easily implemented in interval arithmetic) to determine the maximum of $|\widehat{f''}(t)|$ for $t\leq 2$.

This should all work, but it can't be considered straightforward, and it's a little more computationally intensive than one might in principle wish. Does anybody have a good alternative?

(One possibility is to use a more rapidly converging approximation to erf or to $F$ than the one given by a Taylor series. There is a continued fraction approximation to $F$, for instance; there are some papers from the early 1970s about this. However, I can't help wondering whether there is a cleaner method altogether.)

• It might make more sense to write your method as part of the question itself. (use some sort of spacing or lines to separate it from the question proper) – Chill2Macht Jun 18 '16 at 12:54
• All right, I'll add it. (I had to make a pause anyhow - was in a lecture.) – H A Helfgott Jun 18 '16 at 13:39