Fourier transform of vanishing function in $L^2$

Question

Let $u$ be a function in $L^{2}(\mathbb{R}^{3})$,

Question1:

Is $\left \| u \right \|_{L^{2}} =0 $ implies that $F\left(u\right)=0$?

Question2:

Is $\left \| u \right \|_{L^{2}} > 0$ implies that there exists a positive constant $c> 0$ such that $F\left ( u \right )> c$?

Note that $F(u)$ is the Fourier transform of $u$.

Answer 1

The answer for question 1 is yes, because the Fourier transform $\mathcal{F}:L^2\to L^2$ is an isomorphism between Hilbert spaces. In particular for linearity it maps 0 to 0 (actually you only need $\mathcal{F}$ to be a linear mapping, which is easy to verify).

Remark that the $0$ of $L^2$ is a class of equivalence modulo a.e. equality, so you would have $\mathcal{F}(u)=0$ a.e.

The second part is false. To see it, take any $u\in L^2, u\ne0$ (and hence with strictly positive norm) and consider $\mathcal{F}(u)$. If there is a constant $C$ for which $\mathcal{F}(u)>C$, then $\mathcal{F}(-u)=-\mathcal{F}(u)<-C<0$ and $-u$ and $u$ have the same norm.

Answer 2

If $u \neq 0$ on a subest of $\mathbb{R}^{3}$ then $F(u) \neq 0$ and thus $/F(u)/ \neq 0$ Therefore, $\left \| u \right \|_{L^{2}} $ = $\left \| F(u) \right \|_{L^{2}} $$\neq 0$. the inverse is true.