Weak solution to the free particle Schrodinger equation I'm trying to prove that $\forall\psi_0\in L^2(\mathbb{R})$,the integral
$$\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ikx}\left[\hat{\psi_0}(k)e^{-i\omega(k)t}\right]dk$$ satisfies the Schrodinger equation in the weak sense, namely, $$\int_{\mathbb{R}^2}\psi(x,t)\left[\frac{\partial\chi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^2\chi}{\partial x^2}\right]dxdt=0$$ for any $\chi$ smooth and compactly supported.
My idea is that I can prove this for functions $\psi_0$ in the space $\phi(\mathbb{R})$ of smooth and rapidly decreasing functions (the Schwartz space) and extend it to all $L^2$ functions. Indeed, the proof for Schwartz functions is easy since they are decreasing fast enough so that $k^2\hat{\psi_0}$ is integrable, enabling us to apply the dominated convergence theorem and integration by part. 
Now the Schwartz space $\phi(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ and thus $\forall\psi_0\in L^2(\mathbb{R})$, $\exists(\psi_{0,n})\subseteq\phi(\mathbb{R})$ such that $\psi_{0,n}\rightarrow\psi_0$ in $L^2$ norm. However, I'm not sure in what sense $\psi_n(x,t)$, defined similar to the first integration, converges to $\psi(x,t)$, defined in the first equation. Can I say that $\psi_n(x,t)$ converges to $\psi(x,t)$ in the sense of $L^2(\mathbb{R}^2)$? Why?
By the way, it would be equally appreciated if one can give another proof, for example, a direct proof without utilizing the Schwartz space. Thanks for your time and patience in advance!
 A: First observe that in general the function $\psi(x,t)$, defined by you as a general weak solution of the free Schroedinger equation, doesn't belong to $L^2(\mathbb R^2)$. Indeed, we usually have $\int_{\mathbb R}\left|\psi(x,t)\right|^2dx=\left\|\psi_0\right\|^2_{L^2(\mathbb R)}$ $\forall t\in \mathbb R$,  and then $\int_{\mathbb R^2}\left|\psi(x,t)\right|^2dx\,dt=\infty$ by the Fubini-Tonelli's theorem. We cannot hope that such functions could belong to $L^2(\mathbb R^2)$. At least you have to consider a finite time interval. Otherwise the function $\psi(x,t)$ could be seen as an element of the space $L^{\infty}(L^2(\mathbb R))$, i.e. the space of the function $f:t\in\mathbb R\rightarrow f(t)\in L^2(\mathbb R)$ with bounded sup-norm.
Nevertheless the function $\psi(x,t)$ is a weak solution of the free Schroedinger equations for every $\psi_0\in L^2(\mathbb R)$. Since $H_0=-\frac{\hbar^2}{2m}\Delta$ is a self-adjoint operator in $L^2(\mathbb R)$, by common results of the semi-group/group theory we have that $e^{-iH_0t/\hbar}$ is a unitary operators in $L^2(\mathbb R)$, and by varying  $t$ in $\mathbb R$ they also form a strongly continuos group of operators. Then (see Engel - Nagel, One-Parameter Semigroups for Linear Evolution Equations, Lemma II.1.3 (iii)-(iv) for example) $\int_0^te^{-iH_0s/\hbar}\psi_0\, ds$ is in the domain of $H_0$ for every $t\in\mathbb R$ and
$$ e^{-iH_0t/\hbar}\psi_0=-\frac{i}{\hbar} H_0\int_0^te^{-iH_0s/\hbar}\psi_0\, ds+\psi_0$$
for every $\psi_0\in L^2(\mathbb R)$.
Defined $\psi(t)=e^{-iH_0t/\hbar}\psi_0$, then $\psi(t)$ is a "mild solution" of the Schroedinger equation, i.e.
$$\psi(t)=-\frac{i}{\hbar}H_0\int_0^t \psi(s)ds+\psi_0$$
which correspond to the usual Schroedinger equation only when $\psi_0$ is in the domain of $H_0$ (in that case the operator $H_0$ can be moved inside the integral, and by time-differentiation we get such equation).
We now have to check that (i) $\psi(t)$ corresponds to the function $\psi(x,t)$ given by you and that (ii) a mild solution is a weak solution too. 
Defined with $U_F$ the Fourier's transform operator, which is a unitary operator in $L^2(\mathbb R)$, by the spectral theorem we have
$$ \psi(t)=U_F^{-1}\left[\left(U_F e^{-iH_0t/\hbar}U_F^{-1}\right)\left(U_F\psi_0\right)\right]=U_F^{-1}\left[e^{i\hbar k^2t/2m}\hat{\psi}_0(k)\right], \quad k\in\mathbb R$$
that corresponds to your expression of $\psi(x,t)$ with $\omega(k)=-\hbar  k^2/2m$ when $\hat{\psi}_0\in L^1(\mathbb R)\cap L^2(\mathbb R)$ (as you probably know your expression is indeed "formal", since the integral $$\int_{-\infty}^{\infty} e^{ikx}\left[\hat{\psi}_0(k)e^{-i\omega(k)t}\right]dt$$ is defined only when the integrand function is absolutely integrable). 
Let us to prove the second assertion. Given $\chi(x,t)$ a smooth real function with compact support, then for every $t\in \mathbb R\,$ it results that $\chi(t):=\chi(\cdot,t)$ and $\dot{\chi}(t):=\partial \chi(\cdot,t)/\partial t$ are both in $L^2(\mathbb R)$ and in the domain of $H_0$ too. Denoted by $\left<\cdot,\cdot\right>$ the inner product in $L^2(\mathbb R)$, i.e. $$\left<\alpha,\beta\right>=\int_{\mathbb R}\alpha(x)^*\beta(x)dx,\quad \alpha, \beta\in L^2(\mathbb R)$$ from the above definition of mild solution we get
$$\int_{\mathbb R^2}\frac{\partial \chi(x,t)}{\partial t}\psi(x,t)dxdt =\int_{\mathbb R}\left<\dot{\chi}(t),\psi(t)\right>dt=\int_{\mathbb R}\left<\dot{\chi(t)},-\frac{i}{\hbar}H_0\int_0^t e^{-iH_0s/\hbar}\psi_0\,ds\right>dt=\int_{\mathbb R}\left(-\frac{i}{\hbar}\right)\left<H_0\dot{\chi}(t),\int_0^t e^{-iH_0s/\hbar}\psi_0\,ds\right>dt=\int_{\mathbb R}\frac{i}{\hbar}\left<H_0\chi(t), e^{-iH_0t/\hbar}\psi_0\right>dt=\int_{\mathbb R}\frac{i}{\hbar}\left<H_0\chi(t), \psi(t)\right>dt=-\int_{\mathbb R^2}\frac{i\hbar}{2m}\frac{\partial^2\chi(x,t)}{\partial x^2}\psi(x,t)dx\,dt$$ 
where in the forth equality we have commuted $\partial/\partial t$ with $H_0$ being $\chi$ smooth enough to change the order of derivation and then we have applyied an integration by parts.
A: Hint: use the fact that
$$
\|\psi(\cdot,t)-\psi_n(\cdot,t)\|_{L^2(\mathbb{R})}
=\|\hat{\phi}_0-\hat{\phi}_{0,n}\|_{L^2(\mathbb{R})}.
$$
