Inequality with integral and max of derivative

[RaamGeneral]

Hi. I was reading Lighthill, Introduction to Fourier Analysis and Generalised Functions and in page 17 there is an example/proof where I can't make sense of the following step:
$$
\left| \int_{-\infty}^{+\infty} f_n(x)(g(x)-g(0)) \, \mathrm{d}x \right| \le
\max{ \left| g'(x) \right| } \int_{-\infty}^{+\infty} f_n(x)\left| x \right| \, \mathrm{d}x
$$

where in particular
$$
f_n(x)=\sqrt{\frac{n}{\pi}} \mathrm{e}^{-n x^2}
$$

I have actually tried for some time, exploring some inequalities like Cauchy–Schwarz.

Also, I couldn't get the preview of this post to work, while creating, is this a known issue?

 

[anuttarasammyak]

It seems
$$|g(x)-g(0)|=|\int_0^x g'(t) dt| < max |g'| |x|$$
where max |g'| is maximum in domain 0<t<x .

 

[pasmith]

First use $$\left| \int_{-\infty}^{\infty} f(x)\,dx \right| \leq \int_{-\infty}^\infty |f(x)|\,dx.$$ The given result would then follow from
$$|g(x) - g(0)| = \left|\int_0^x g'(t)\,dt\right| \leq |x|\max |g'|.$$

 

[anuttarasammyak]

For the given f_n(x), the integral is
$$\int_{-\infty}^{+\infty} f_n(x)|x|dx=\int_0^{+\infty}\sqrt{\frac{n}{\pi}} \mathrm{e}^{-n t} dt=(n\pi)^{-1/2}$$