It may be an idiot question but it really has bothered me for a long time.
For two distributions (generalized functions) $u,v \in \mathcal{D}'( \mathbb{R}^n)$, if $u=v$ in the distributional sense (i.e. for any test function $\phi \in \mathcal{D}(\mathbb{R}^n)=C_c^\infty(\mathbb{R}^n),$ we have $(u,\phi)_{\mathcal{D}',\mathcal{D}}=(v,\phi)_{\mathcal{D}',\mathcal{D}}.$) Now if we know that $v$ is also in $L^2(\mathbb{R}^n)$ (square integrable function) ,can we deduce that $u$ is also in $L^2(\mathbb{R}^n)?$
First, if I know $v$ is a distribution and additionally $u\in L^2(\mathbb{R}^n),$ for any test function $\phi \in \mathcal{D}(\mathbb{R}^n)$, is $(v,\phi)_{\mathcal{D}',\mathcal{D}}$ must be written as $\int_{\mathbb{R}^n} v(x)\phi(x)dx$? Besides, how can I deduce the property of distribution $u$?