Density argument in the definition of weak solution of PDEs

Question

I don't remember how to properly prove the following. Let us consider the following PDE (transport equation): $$ (1) \quad \left\{\begin{array}{l} u_t+u_x=0, \\ u(t,0)=h(t), \\ u(0,x)=u_0(x). \end{array}\right. $$ where $t \in (0,T)$ and $x \in (0,1)$, and $h \in L^2_{\mathrm{loc}}(0,+\infty)$ and $u_0 \in L^2(0,1)$. Then, the classical way to define the notion of weak solution is to multiply (1) by $\varphi \in C^1([0,T]\times[0,1])$ and do integrations by parts. This leads to the following definition: $u$ is a solution to (1) if $u \in C^0([0,+\infty);L^2(0,1))$ and satisfies, for every $T>0$, $$ (2) \quad \int_0^1 u(T,x)\varphi(T,x) \, dx - \int_0^1 u_0(x)\varphi(0,x) \, dx +\int_0^T \int_0^1 u(t,x) \left(-\varphi_t(t,x)-\varphi_x(t,x)\right)\, dx dt -\int_0^T h(t) \varphi(t,0) \, dt=0, $$ for every $\varphi \in D$, where $$D=\left\{\varphi \in C^1([0,T]\times[0,1]), \quad \varphi(t,1)=0, \quad \forall t \in [0,T]\right\}.$$ Now, we see that each term makes sense in (2) if $\varphi \in E$, where $$E=\left\{\varphi \in H^1(0,T;L^2(0,1)) \cap L^2(0,T;H^1(0,1)), \quad \varphi(t,1)=0 \mbox{ a.e. } t \in (0,T)\right\}.$$ My question is: How to prove that a function $u \in C^0([0,+\infty);L^2(0,1))$ is a solution to (1) if, and only if, it satisfies (2) for every $\varphi \in E$ (and every $T>0$) ?