Inequality constant in Papa Rudin

Question

The following inequality appears in Rudin's Real and Complex Analysis, 3ed, in the proof of ($f$) in Theorem 9.2 (Fourier Transforms): if $x\in\mathbb{R}$ and $\phi(x,u):= (e^{-ixu} - 1)/u$ then $|\phi(x,u)|\le |x|$ for all real $u\ne 0$. It seems to me that this is incorrect and that the inequality should read "$\le 2|x|$". Can anyone please clarify?

Answer 1

If $x\neq0$ then we have the following: $$\begin{align}\lvert\phi(x,u)\rvert&=\left|\frac{e^{-ixu}-1}{u}\right|=\frac{\lvert e^{-ixu}-1\rvert}{\lvert u\rvert}\\ &=\frac{\lvert 1-e^{ixu}\rvert}{\lvert u\rvert}=\frac{\lvert(1-\cos(xu))-i\sin(xu)\rvert}{\lvert u\rvert}\\ &=\frac{\sqrt{(1-\cos(xu))^{2}+\sin^{2}(xu)}}{\lvert u\rvert}=\frac{\sqrt{2-2\cos(xu)}}{\lvert u\rvert}\\ &=\sqrt{2}\frac{\sqrt{1-\cos(xu)}}{\lvert u\rvert}=2\frac{\sqrt{\frac{1-\cos(xu)}{2}}}{\lvert u\rvert}=2\frac{\lvert\sin(\frac{xu}{2})\rvert}{\lvert u\rvert}\\&=\lvert x\rvert\frac{\lvert\sin(\frac{xu}{2})\rvert}{\lvert\frac{xu}{2}\rvert}\le\lvert x\rvert \end{align}$$.

If $x=0$ then $\phi(0,u)=0$ so $\vert\phi(0,u)\rvert\le0$. In both cases we have $\lvert\phi(x,u)\rvert\le\lvert x\rvert$.

Answer 2

Here is a one liner:

$$ \left| \frac{e^{-ixu}-1}{u} \right|=\left| \int_0^x e^{-itu}dt \right|\le\int_0^{|x|} |e^{-itu}|dt=|x|.$$