$L^\infty (\Bbb R^n)$ bound for an inverse Fourier transform.

Question

Let $\phi:\Bbb R^n\to\Bbb R$ be a smooth and compactly supported function ($\phi\equiv1$ on some ball around $0$). I want to show that the function $f:\Bbb R^n\to\Bbb C$ defined by $$ f(x) := \int_{\Bbb R^n} \phi(\xi)\frac{\xi_j}{|\xi|^2}e^{2\pi i x\cdot\xi} \,d\xi $$ for a fixed $j\in\{1,\dots,n\}$is unformly bounded, i.e. $f\in L^\infty$. Here $|\xi|:=(\xi_1^2+\cdots+\xi_n^2)^{1/2}$.

I get this expression from a lecture note on harmonic analysis but I don't know how to prove it. Could anyone please suggest how to show this?

Answer 1

We have that $$|f(x)| \leq \int_\mathbb{R^n} \bigg|\phi(\xi)\frac{\xi_j}{|\xi|^2}e^{2\pi i x\cdot\xi}\bigg| \,d\xi \leq \int_{\mathbb{R}^n} \frac{|\phi(\xi)|}{|\xi|} \, d\xi$$ since $|\xi_j| \leq |\xi|$ and $|e^{2\pi i x \cdot \xi}| = 1$.

Since the right hand side is independent of $x$, we only need to check that the integral is finite.

Now $\phi$ is smooth and compactly supported so we only have trouble on the right hand side near to the origin. So it suffices to consider $$\int_{B(\varepsilon)} \frac{|\phi(\xi)|}{|\xi|} \, d\xi$$ for small $\varepsilon >0$. $\phi$ is bounded and for $n>1$, $\frac{1}{|x|}$ is integrable on $B(\varepsilon)$ and hence this integral is finite.

In the case $n=1$, as far as I can tell the integral defining $f$ doesn't make sense since close to the origin the integrand has real part behaving like $\frac{cos(2 \pi \xi)}{\xi}$ (say for $x=1$) which is not integrable on a neighbourhood of $0$.

Answer 2

In the case $n=1$ the $p.v.$ definition seems to work. Fix a small $k>0$. Then for $x\ne 0$ $$p.v.\int_{\mathbb R} \phi(\xi)\xi^{-1}e^{2i\xi x}\,d\xi= \lim_{\delta\to 0}\left(\int_{\delta\le|\xi|\le \delta^{-1}} \xi^{-1}e^{2i\xi x}\,d\xi+\int_{k\le|\xi|}\phi(\xi)\xi^{-1}e^{2i\xi x}\,d\xi-\int_{ k\le |\xi|\le \delta^{-1}}\xi^{-1}e^{2i\xi x}\,d\xi\right).$$ Then, as $\lim_{\delta\to 0}\int_{\delta+c}^{\delta^{-1}}\sin(\xi)/\xi\,d\xi$ exists for every $c\ge0$ $$p.v.\int_{\mathbb R} \phi(\xi)\xi^{-1}e^{2i\xi x}\,d\xi= i\int_{\mathbb R}\xi^{-1}\sin(\xi)\,d\xi+C(x,k)-i\int_{ |x|2k\le |\xi|< \infty}\xi^{-1}\sin(\xi)\,d\xi.$$ Now observe that $\sup_x|C(x,k)|<\infty$ and $\sup _x |\int_{ |x|2k\le |\xi|< \infty}\xi^{-1}\sin(\xi)\,d\xi|<\infty$ as $x\mapsto\int_x^\infty\xi^{-1}\sin(\xi)\,d\xi$ is a continuous bounded function on $[0,\infty)$. For $x=0$ the third term is $0$.