Is a mild solution the same thing as a weak solution? In the book on PDEs by L. Evans, a solution to the heat equation with Dirichlet boundary conditions: 
$$\tag{HP}
\begin{cases}\displaystyle
\frac{\partial u}{\partial t}=\Delta u & x\in U,\ t\in(0, T)\\
u=0 & \text{on } \partial U\\
u=g\in L^2(U) & \text{at time }t=0
\end{cases}
$$
is constructed by means of the Galerkin method. This happens to be a weak solution in the sense that 
$$
u\in L^2(0, T; H^1_0(U)),\quad u'\in L^2(0, T; H^{-1}(U))
$$
(the time derivative is taken in distributional sense $^{[1]}$) and it satisfies the equation where the Laplacian is taken in the "weak sense" of elliptic theory $^{[2]}$:
$$\tag{1}
\int_U \frac{\partial u}{\partial t}(x, t)v(x)\, dx = -\int_U \nabla u (x, t)\cdot\nabla v(x)\, dx,\quad \forall v \in H^1_0(U).
$$
(see §7.1 of the second edition).
Later in the same book, (HP) is treated by means of the semigroup approach. Namely, it is proved that the unbounded operator $A=\Delta$ defined on the domain $D(A)=H^2(U)\cap H^1_0(U)$ generates a contraction semigroup $S_t$ on $L^2(U)$ space. Therefore the function 
$$\tag{2}u(x, t)=(S_tg)(x)$$
is a strong solution to (HP) if $g\in D(A)$. (see §7.4).

Question.  Formula (2) makes sense even if $g\notin D(A)$. Is it true that in this case the function $u$ defined by (2) is a weak solution in the sense of formula (1)?

In the book Vector-valued Laplace transforms and Cauchy problems by V.A., 2nd edition $^{[3]}$, this kind of solutions are called mild solutions (Definition 3.1.1 - see also Proposition 3.1.9). This explains the title.

Notes.
$^{[1]}$ Explicitly, 
$$-\int_0^T u(t)\phi'(t)\, dt= \int_0^t u'(t)\phi(t)\, dt,\quad \forall \phi\in C^{\infty}_c(0, T).$$
This is somewhat related to this other question on vector-valued distributions. 
$^{[2]}$ Also called energetic extension of the Laplacian. 
$^{[3]}$ Recommended to me some time ago by the user lvb in this great answer.
 A: The easiest way to show the coincidence of the two solutions is to verify that
both are weak solutions of the class $L^2(0, T; H^1_0(U))$. For problem (HP),
a function $u\colon\, (0,T)\to H^1_0(U)$ is called a weak solution of the class
$u\in L^2(0, T; H^1_0(U))$ if $u$ satisfies the integral identity
\begin{align*}
-\!\!\int\limits_{Q_T}\! uv_t\,dxdt+\int\limits_{Q_T} \nabla u\cdot\nabla v\,dxdt=
\!\int\limits_U g(x)v(x,0)\,dx\tag{$\ast$}\\
\forall\, v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0,
\end{align*}
where $Q_T=U\times (0,T)$. For the linear problem (HP), uniqueness of the weak solution $(\ast)$ is rather obvious. Indeed, take $g=0$ and notice that the choice of a test function
$$
v(x,t)=\int\limits_t^T u(x,s)\,ds
$$
implies that
$$
\int\limits_{Q_T}\! |u(x,t)|^2\,dxdt+\frac{1}{2}\int\limits_{U}
\Bigl|\int\limits_0^T\nabla u(x,s)\,ds\Bigr|^2dx=0,
$$
i.e., $u=0$ a.e. in $Q_T\,$.
To verify that a weak solution $(1)$ will be a weak solution $(\ast)$, notice that the weak solution $u$ in the sense $(1)$ satisfies the integral identity
$$
\int\limits_{Q_T} u_t v \, dxdt 
+\int\limits_{Q_T} \nabla u \cdot\nabla v\, dxdt=0 \tag{$\ast\ast$}
$$
for all test functions $v=v(x,t)$ stepwise w.r.t. variable $t$, hence for all 
$v\in L^2(0, T; H^1_0(U))$, and hence for all 
$$
v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0.
$$ 
Integrating by parts in $(\ast\ast)$ results in the identity $(\ast)$.
To verify that the mild solution will be a weak solution $(\ast)$ multiply
the equation
$$
\Delta\!\!\int\limits_0^t u(x,s)\,ds=u(x,t)-g(x)
$$
by $v_t(x,t)$ with $v\in H^1(Q_T)\colon\,v|_{\partial U}=0,\;v(x,T)=0$, and
then integrate over $Q_T$ to find
$$
\int\limits_{Q_T}\Bigl(\Delta\!\int\limits_0^t u(x,s)\,ds\Bigr)v_t(x,t)\,dxdt=
\int\limits_{Q_T}uv_tdxdt+\int\limits_{U}g(x)v(x,0)\,dx,
$$
where the left hand side equals
\begin{multline*}
\int\limits_{U}dx\int\limits_0^T v_t(x,t)\,dt\int\limits_0^t\Delta u(x,s)\,ds=
\int\limits_{U}dx\int\limits_0^T\Delta u(x,s)\,ds\int\limits_s^T v_t(x,t)\,dt\\
=-\int\limits_{Q_T}\Delta u(x,s)v(x,s)\,dxds=
\int\limits_{Q_T}\nabla u(x,s)\cdot\nabla v(x,s)\,dxds
\end{multline*}
which results in the identity $(\ast)$.
