The below picture comes from "The Cauchy-Schwarz Master Class". I am interested in the argument in the red box. Note that the two circles indicate a typo by the author. They should be replaced by $x+h$ and $h$ respectively.
My question is that how does one arrive at the last inequality in the red box? I've tried to integrate it out and apply the inequalty $h^{1/2}\leq f(x)/D(x)$ but this doesn't work.
Best Answer
The author uses 7.13 together with the conditions on $h$ above the box:
$$ |f(x+t)| \geq |f(x)| -\sqrt t D(x)$$
and
$$0\leq h \leq \frac{f(x)}{D(x)} \ \Rightarrow f(x) \geq hD(x) \geq 0 \Rightarrow |f(x)| = f(x)$$
Hence,
$$\int_0^{h}(x+t)^2|f(x+t)|^2dt \geq \int_0^{h}(x+t)^2||f(x)| -\sqrt t D(x)|^2dt $$ $$\stackrel{f(x)\geq 0}{=}\int_0^{h}(x+t)^2|f(x) -\sqrt t D(x)|^2dt$$ $$\stackrel{f(x) \geq hD(x) \geq 0}{\geq}hx^2\left(f(x)-\sqrt h D(x)\right)^2$$
The last inequality hold because for $t\in [0,h]$ you have
$$(x+t)^2|f(x) -\sqrt t D(x)|^2 \stackrel{f(x) \geq hD(x) \geq 0}{\geq} x^2\left(f(x)-\sqrt h D(x)\right)^2$$
This is independent of $t$ - the variable of integration. Hence, the factor $h$ in the integral inequality.