Let $J = (0,T)$, $T > 0$, $G = (a,b) \subset \mathbb{R}$ (finite interval), and $f \in C(J;L^2(G))$. I consider the heat equation with zero Dirichlet boundary conditions and with initial value $u_0 \in L^2(G)$, \begin{align} \begin{cases} \frac{\partial u(t,x)}{\partial t} - \frac{\partial^2 u(t,x)}{\partial x^2} &= f(t,x) \quad \text{in } J \times G, \\ u(t,x) &= 0 \quad \text{on } J \times \partial G, \\ u(0,x) &= u_0(x) \quad \text{in } G. \end{cases} \end{align}
I am given the following problem: Show that if $u$ is a smooth solution of this problem, then for all $v \in H^1_0(G)$, there holds $$ \frac{d}{dt} ( u(t),v )_{L^2(G)} + a(u(t), v) = (f(t), v)_{L^2(G)}, \quad \forall t \in J, $$ where $$ a(\phi, \psi) = \int_G \frac{\partial \phi(x)}{\partial x} \frac{\partial \psi(x)}{\partial x} \, dx, \quad (\phi, \psi)_{L^2(G)} = \int_G \phi(x) \psi(x) \, dx. $$
My attempt: Let $u$ be a smooth solution and let $v \in H^1_0(G)$ (this is the Sobolev space for $p = 2$ such that the continuous modification of $v$ vanishes on the boundary of $G$) and $t \in J$.
If I simply multiply the heat equation by $v$ and integrate over $G$, I have (and since $u$ is smooth, I can take out the time derivative) $$ \frac d{dt}\int_G u(t,x)v(x) dx - \int_G \partial_{xx}u(t,x) v(x) dx = \int_Gf(t,x)v(x)dx. $$ A simple partial integration for the middle term would yield the result. However, I do not know why I can use partial integration. Since $v \in H^1_0(G)$, I know that for any test function $\phi \in C^1_0(a,b)$ (i.e. continuously differentiable functions which vanish on the boundary of $G$) I have $$ \int_G v \phi' dx = - \int_G v' \phi dx. $$ To apply this, I would have to identify $\partial_{xx}u(t,x)$ as the derivative of a test function, which I am not able to (especially because of the boundary condition that it must vanish at $\partial G$).
Where is my mistake? Or should $v$ be in $C^1_0(G)$ instead of $H^1_0(G)$ and there is a mistake in the task?
Thanks in advance!
Actually it suffices to use integration by parts and use the fact that $v\in H^1_0(G)=\{u\in L^2(G):\exists g\in L^2(G),\int_Gu\phi'=\int_G-g\phi,\;\forall\phi\in C^1_0(G),\exists\tilde{u}\in C^0(G),\,u=\tilde{u}\, \text{a.e. on }G,\, \,\tilde{u}\vert_{\partial G}=0 \}$ where $\tilde{u}$. is the unique continuous modification of $u$. Then $-\int_G\partial_{xx}uvdx=-\partial_xuv\vert_{\partial G}+\int_G\partial_xu\partial_xvdx$ and the second term is zero due to the boundary conditions imposed on $v$.