First we look for a distributional solution. Remember that, as an distribution, $\Delta u$ is defined by $$\langle\Delta u,v \rangle=-\langle\nabla u,\nabla v\rangle,\ \forall\ v\in C_0^\infty(\Omega). \tag{1}$$
From $(1)$, we can say that a solution in the distributional sense, is a function $u\in H^1(\Omega)$ with $Tu=g$ satisfying $$\int_\Omega \nabla u\nabla v=0,\ \forall\ v\in C_0^\infty(\Omega). \tag{2}$$
By density we may conclude from $(2)$ that $$\int_\Omega \nabla u\nabla v=0,\ \forall v\in H_0^1(\Omega). \tag{3}$$
This is not the only weak formulation for this problem, however, it is the one which comes from a variational problem, to wit, let $F:\{u\in H^1(\Omega):\ Tu=g \}\to \mathbb{R}$ be defined by $$Fu=\frac{1}{2}\int_\Omega |\nabla u|^2.$$
Note that $(3)$ can be rewritten as $$\langle F'(u),v\rangle=0,\ \forall\ v\in H_0^1(\Omega). \tag{4}$$
Also note that, once $\{u\in H^1(\Omega):\ Tu=g \}$ is a closed convex set of $H^1(\Omega)$ and $F$ is a coercive, weakly lower semi continuous function, we have that $F$ has a unique global minimum which satisfies $(4)$.
By not considering any kind of derivative of $u$, you can also use another weak formulation: let $C_0^{1,\Delta}(\overline{\Omega})=\{u\in C_0^1(\overline{\Omega}):\ \Delta u \in L^\infty(\Omega)\}$. A "very" weak solution, is a function $u\in L^1(\Omega)$ satisfying $$\int_\Omega u\Delta v=-\int_{\partial\Omega}g\frac{\partial v}{\partial \nu },\ \forall\ v\in C_0^{1,\Delta}(\overline{\Omega}).$$
In your setting, I mean, when $g\in H^{1/2}(\Omega)$, it can be proved that both definitions are equivalent. For references, take a look in the paper Elliptic Equations Involving Measures from Veron. It has a PDF version here. Take a look in page 8.
To conclude, I would like to adress @JLA, which gave a comment in OP; in the end, what we really want is a $H^2$ function (or more regular), because we are working with the Laplacean and it is natural to have two derivatives.
It can be proved, by using regularity theory, that $u$ is in fact in $H^2$, however, there is a huge difference between proving that $u$ is in $H^2$ after finding it in $H^1$ by the above methods and finding directly $u\in H^2(\Omega)$ by another method. Note, for example, that none of the methods above, does apply if we change $H_0^1(\Omega)$ by $H^2(\Omega)$.
Is the above correct?
I'd say yes, provided that $u\in L^2(0,T;L^2(U))$ instead of $L^2(0,T;H^{-1}(U))$. To see this, we could use the integration by parts formula:
Theorem (Dautray, p. 473 and p. 477). Let $V$ and $H$ be Hilbert spaces such that
- $V$ is continuously and densely embedded into $H$
- $H$ is identified with $H'$
- $H'$ is continuously and densely embedded into $V'$
(A) If $u\in L^2(a,b;V)$ and $u_t\in L^2(a,b;V')$, then $u\in C([a,b]; H)$.
(B) If $u,v\in L^2(a,b;V)$ and $u_t,v_t\in L^2(a,b;V')$, then
$$\int_a^b\langle u_t(t),v(t)\rangle_{V'\times V}\;dt=(u(b),v(b))_H-(u(a),v(a))_H-\int_a^b\langle u(t),v_t(t)\rangle_{V\times V'}\;dt\tag{1}$$
which has a sense by (A).
Therefore:
If $u,v\in L^2(0,T;H_0^1(U))$ and $u_t,v_t\in L^2(0,T;H^{-1}(U))$, then
\begin{align*}
\int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=(u(T),v(T))_{L^2}-(u(0),v(0))_{L^2}\\
-\int_0^T\langle u(t),v_t(t)\rangle_{H_0^1\times H^{-1}}\;dt\tag{2}
\end{align*}
If we have the additional regularity $v\in C_c^\infty(0,T;H_0^1(U))\subset L^2(0,T;H_0^1(U))$, then
$v_t\in C_c^\infty(0,T;H_0^1(U))\subset C_c^\infty(0,T;H^{-1}(U))\subset L^2(0,T;H^{-1}(U))$. In this case, the last equality becomes
$$\int_0^T\langle u_t(t),v(t)\rangle_{H^{-1}\times H_0^1}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt.\tag{3}$$
If we have the additional regularity $u_t\in L^2(0,T;L^2(U)) \subset L^2(0,T;H^{-1}(U))$, then the last equality becomes
$$\int_0^T(u_t(t),v(t))_{L^2}\;dt=-\int_0^T( u(t),v_t(t))_{L^2}\;dt,\tag{4}$$
which can be rewritten as
$$\int_0^T\int_U u_t(t,x)v(t,x)\;dx\;dt=-\int_0^T\int_U u(t,x)v_t(t,x)\;dx\;dt.\tag{5}$$
Thus $(5)$ is the same as $(4)$, which is a particular case of $(3)$, which is a particular case of $(2)$, which is a particular case of $(1)$.
Best Answer
If $u$ has ordinary second order derivatives and $\phi$ is a test function, then $$\langle T_{-\Delta u}, \phi \rangle = \int_\Omega (-\Delta u) \phi = \int_\Omega Du \cdot D\phi.$$ The last integral makes sense even if $u$ belongs only to $H^1(\Omega)$, and the distribution $T$ defined by $$\langle T,\phi \rangle = \int_\Omega Du \cdot D\phi$$ is called the distributional Laplacian of $u$. If it happens that $$\int_\Omega Du \cdot D\phi = \int_\Omega f\phi$$ for all test functions $\phi$ then you have $T = T_f$. This is what is meant by $f = -\Delta u$ in the distributional sense.
Your second question requires some regularity theory to ensure that $u$ has second order weak derivatives.