Proving that a function related to the Fourier transform of the unit ball is in $L^1(\mathbb{R}^3)$

Question

Let $B = \{x \in \mathbb{R^3}\ : ||x|| \leq 1\}$ be a unit ball in $\mathbb{R^3}$. Having computed the Fourier Transform of $B$'s characteristic function $\hat{f}(k)$, where:

$\hat{f}(k) = \int_{B} e^{-2 \pi i \langle k,x \rangle}dx, k \in \mathbb{R^3}$, I have difficulty proving that $\frac{\hat{f}(k)}{||k||^2} \in L^1(\mathbb{R^3})$.

I know that $|\hat{f}(k)| \leq \hat{f}(0)$, so $\frac{\hat{f}(k)}{||k||^2} \leq \frac{\hat{f}(0)}{||k||^2}$, and $\frac{\hat{f}(0)}{||k||^2}$ is integrable on any ball. But I don't think that thats enough to conclude that $\frac{\hat{f}(k)}{||k||^2}$ is integrable on any ball. Concerning integration on the outside of the ball, I am unsure how to proceed. Is there a general method of dealing with these types of problems?

Answer 1

You have correctly argued that $$ \int_{\|k\|\le 1} \frac{|\hat f (k)|}{\|k\|^2} \,dk < \infty $$ A bit more detail: a measurable function bounded by an integrable function is itself integrable. Loosely speaking, anything defined by an explicit formula is going to be Lebesgue integrable. To have a precise proof, note that $\hat f$ is continuous (since it's the Fourier transform of an $L^1$ function). Therefore, $\hat f(k)/\|k\|^2$ is continuous on $\mathbb{R}^n\setminus \{0\}$, hence is measurable there. Adding one point to the domain does not change measurability, since one point is a set of measure zero.

On the exterior of the ball, the Cauchy-Schwarz inequality does the job as both $\hat f$ and $1/\|k\|^2$ are in $L^2$ on that domain: $$ \int_{\|k\|\ge 1} \frac{|\hat{f}(k)|}{\|k\|^2} \,dk \le \left( \int_{\|k\|\ge 1} |\hat{f}(k)|^2 \,dk \right)^{1/2} \left( \int_{\|k\|\ge 1} \frac{1}{\|k\|^4} \,dk \right)^{1/2} <\infty $$

Answer 2

A Simple Bound

To bound the integral, let $f(x)=\big[\|x\|\le1\big]$. Then $$ \begin{align} \left\|\,\lower{1pt}{\hat{f}}\,\right\|_\infty &=\left\|\,f\,\right\|_1\\ &=\frac{4\pi}3\tag1 \end{align} $$ Normally $\left\|\,\lower{1pt}{\hat{f}}\,\right\|_\infty \le\left\|\,f\,\right\|_1$, but $\hat{f}(0)=\int_{\mathbb{R}^3}f(x)\,\mathrm{d}x=\frac{4\pi}3$. Furthermore, Plancherel says $$ \begin{align} \left\|\,\lower{1pt}{\hat{f}}\,\right\|_2 &=\left\|\,f\,\right\|_2\\ &=\sqrt{\frac{4\pi}3}\tag2 \end{align} $$ Hölder says $\int_Xf(x)g(x)\,\mathrm{d}x\le\|f\|_\infty\|g\|_1$; therefore, $$ \begin{align} \int_{\|x\|\le1}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x &\le\left\|\,\hat{f}\,\right\|_\infty\int_{\|x\|\le1}\frac1{\|x\|^2}\,\mathrm{d}x\\ &=\frac{4\pi}3\cdot4\pi\tag3 \end{align} $$ Hölder says $\int_Xf(x)g(x)\,\mathrm{d}x\le\|f\|_2\|g\|_2$; therefore, $$ \begin{align} \int_{\|x\|\gt1}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x &\le\left\|\,\lower{1pt}{\hat{f}}\,\right\|_2\left(\int_{\|x\|\gt1}\frac1{\|x\|^4}\,\mathrm{d}x\right)^{1/2}\\ &=\sqrt{\frac{4\pi}3}\cdot\sqrt{4\pi}\tag4 \end{align} $$ Thus, $$ \begin{align} \int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x &\le\frac{16\pi^2}3+\frac{4\pi}{\sqrt3}\\ &\approx59.893\tag5 \end{align} $$

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A Better Bound

If we split the cases at $\|x\|=r$, we get the bound $$ \begin{align} \int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x &\le\frac{16\pi^2}3r+\frac{4\pi}{\sqrt3}r^{-1/2}\\ &=\left(\frac{4\pi}{\sqrt3}\right)^{4/3}\left[\left(\frac{4\pi}{\sqrt3}\right)^{2/3}r+\left(\frac{4\pi}{\sqrt3}\right)^{-1/3}r^{-1/2}\right]\tag6 \end{align} $$ Since $x^2+\frac1x\ge\left(\frac{27}4\right)^{1/3}$ and $x^2+\frac1x=\left(\frac{27}4\right)^{1/3}$ when $x=\left(\frac12\right)^{1/3}$, $(6)$ gives the bound $$ \begin{align} \int_{\mathbb{R}^3}\frac{\left|\,\lower{1pt}{\hat{f}(x)}\,\right|}{\|x\|^2}\,\mathrm{d}x &\le\left(192\pi^4\right)^{1/3}\\ &\approx26.544\tag7 \end{align} $$