An everywhere zero function

Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be a continuous function such that $\bigintsss_{\mathbb{R}^3} \left| f(x) \right| \, {\rm d}x$ is real. If for every plane $\mathcal{P}$ of $\mathbb{R}^3$ it holds that $\bigintsss_{\mathcal{P}} f(x) \, {\rm d}s =0$ then prove that $f(x)=0$ forall $x \in \mathbb{R}^3$ .

Solution

Since $\bigintsss_{\mathbb{R}^3} \left| f(x) \right| \, {\rm d}x$ is real, we can define the Fourier transform of $f$ as

$\hat{f}(a) = \int \limits_{\mathbb{R}^3} f(x) e^{-2\pi i ax} \, {\rm d}x$

If $a$ is fixed and $t \in \mathbb{R}$ then the planes defined as

$\mathcal{P} =\left \{ x \in \mathbb{R}^3 \bigg| ax = t \right \}$

are parallel and cover $\mathbb{R}^3$ . Integrating over the planes and then over $t$ we have that

$\begin{align*} \hat{f}(a) = \int \limits_{\mathbb{R}^3} f(x) e^{-2\pi i ax} \, {\rm d}x &= \int \limits_{\mathbb{R}} \int \limits_{xa =t} f(x) e^{-2\pi i t} \, {\rm d} \sigma {\rm d}t \\ &= \int \limits_{\mathbb{R}} e^{-2 \pi i t } \int \limits_{ax =t} f(x) \, {\rm d} \sigma {\rm d}t \\ &= 0 \end{align*}$

Hence forall $a \in \mathbb{R}^3$ we have that $\hat{f}(a)=0$ . Since our function is continuous and its Fourier transform is $0$ it follows that $f=0$ forall $x \in \mathbb{R}^3$ .