Proving that : $ \left|\int_a^b\frac{\widehat{f}(\xi)}{\xi}d\xi\right|\le 4 \|f\|_1$ for all $a,b$ and all $f$

Solution 1:

Since $f$ is odd, $$\hat f(\xi)=2i\int_0^\infty f(t)\sin(t\xi)\,dt.$$

We know there exists $c$ so $$\left|\int_a^b\frac{\sin(\xi)}{\xi}\,d\xi\right|\le c$$for all $a,b$.

Edit: In fact we can take $c=\int_{-\pi/2}^{\pi/2}\frac{\sin(\xi)}{\xi}\,d\xi<4$. See below.

It follows that $$\left|\int_a^b\frac{\sin(\xi t)}{\xi}\,d\xi\right|\le c$$for all $a,b$ and all $t\ne0$. Now the result you want follows (with $c$ in place of $4$) from Fubini:

$$\int_a^b\frac{\hat f(\xi)}{\xi}d\xi=2i\int_{0}^\infty f(t)\int_a^b\frac{\sin(\xi t)}{\xi}\,d\xi dt,$$so $$\left|\int_a^b\frac{\hat f(\xi)}{\xi}d\xi\right|\le2\int_{0}^\infty|f(t)|\left|\int_a^b\frac{\sin(\xi t)}{\xi}\,d\xi\right| dt\le c||f||_1.$$(The $2$ disappeared because $2\int_0^\infty|f|=||f||_1$.)

This is what you asked for, since $c<4$.

Edit:

$\left|\int_a^b\frac{\sin(\xi)}{\xi}\,d\xi\right|\le c< 4$.

It's enough to show that $|\int_0^a|\le c/2< 2$.

Note first that $$\left|\int_0^a\right|\le\int_0^{\pi/2}\quad(a>0).$$ This is clear for $0<a<\pi/2$, and for $a>\pi/2$ it follows by an "alternating series" argument, writing the integral as a sum of integrals over $[k\pi/2,(k+1)\pi/2]$ plus a bit left over at the end. (Or you could clean that up a bit, eliminating the leftover bit at the end, by noting that the critical points of the function $a\mapsto\int_0^a$ are $a=k\pi/2$.)

And $$\int_0^{\pi/2}\frac{\sin(\xi)}{\xi}\,d\xi\le\int_0^1 d\xi+\int_1^{\pi/2}\frac{d\xi}{\xi}=1+\ln(\pi/2)<2,$$since $\pi/2<2<e$.