How to prove that this distribution is $L^2_{loc}$?

Question

Notation: We define as usual $\mathcal{S}(\mathbb{R}^d)$ the space of Schwartz functions and $\mathcal{D}(\mathbb{R}^d)=C^{\infty}_c(\mathbb{R}^d)$. We say that a distribution $u$ (i.e. $u \in \mathcal{D}'(\mathbb{R}^d)$) is $L^2_{loc}$ if there exists $f \in L^2_{loc}$ such that for all $\varphi \in \mathcal{D}(\mathbb{R}^d)$ the following holds: $$<u,\varphi>=\int_{\mathbb{R}^d}f(x)\varphi(x) dx$$ We denote $\mathcal{F}:\mathcal{S}'(\mathbb{R}^d) \to \mathcal{S}'(\mathbb{R}^d)$ the Fourier transform.

Question: What I want to understand is the following: if $u \in \mathcal{S}'(\mathbb{R}^d)$ and $u \in L^2_{loc}$ (as an element of $\mathcal{D}'(\mathbb{R}^d)$) can we conclude that $\mathcal{F}(u) \in L^2_{loc}$?

Easy remark: We recall that if we substitute $L^2_{loc}$ with $L^2$ we obtain positive answer due to Fourier-Plancherel formula.

Answer 1

No, we can't conclude that. The constant function $f \equiv 1$ is a tempered distribution in $L^2_{\text{loc}}$, but its Fourier transform is (up to a multiplicative constant) a dirac delta $\delta$, which doesn't belong to $L^2_{\text{loc}}$.

Answer 2

While $L^2$ is stable by the Fourier transform, $L^2_{loc}$ isn't. To see this, take $f$ to be the constant function $f(x)=1$ for all $x\in\mathbb R$. Then clearly $f$ is locally square integrable, but its Fourier transform isn't (it's the Dirac distribution).