My question is derived from the proof of the equation $\Delta f=f$ which has no nonzero solution in $\mathscr{S}'(\mathbb{R}^n)$. The ideal to solve this equation is to use the Fourier transform. By using Fourier transform, we have $-4\pi^2\left | \xi \right |^2\mathscr{F}f(\xi)=\mathscr{F}f(\xi)$. Hence, the Fourier transform of $f$ must be zero at our first glance. However, I don't know the deep reason in tempered distribution sense.
This question can be generalized as follows:
Suppose $f\in \mathscr{S}'(\mathbb{R}^n)$ and $g\in \mathscr{O}^n_M$, where $$\mathscr{O}^n_M:=\left \{h(x)\in C^\infty(\mathbb{R}^n): \forall \alpha, \exists C(\alpha)>0, N(\alpha)\in\mathbb{N}\cup\{0\}, s.t.\quad |D^\alpha h(x)|\leq C(\alpha)(1+|x|^2)^{N(\alpha)}\right \} $$ is the multiplier of $\mathscr{S}'(\mathbb{R}^n)$. Then the product $``gf"$ is well defined by $<gf, \varphi>=<f, g\varphi>$, $\forall \varphi\in\mathscr{S}(\mathbb{R}^n)$.
My question is:
Fix a $g\in \mathscr{O}^n_M$ and $g\neq 0$, if every $\varphi\in\mathscr{S}(\mathbb{R}^n)$ we have $<gf, \varphi>=0$, then can we make a conclusion that $f=0$ in $\mathscr{S}'(\mathbb{R}^n)$ ?
The Fourier transform of a tempered distribution is a distribution. Let $\hat f$ be the Fourier transform of $f\in\mathcal S'(\Bbb R)$. Your equation leads to $$(1+4\pi|\xi|^2)\hat f =0$$ in the sense of distributions. If we have an equation $gT=0$, where $g\in C^\infty(\Bbb R)$, $T\in D'(\Bbb R)$, then $supp T\subset \{x\in \Bbb R: g(x)=0\}$. In our case $g=(1+4\pi|\xi|^2)$ is never zero, hence $ supp \hat f=\emptyset$, which leads to $\hat f=0$ in the sense of distributions. Fourier transform is injective, hence $f$ is zero, too.