[proofplan] We approximate the evolution equation on finite-dimensional subspaces of $V$ and solve the resulting measurable linear ODE systems. Testing each Galerkin equation against its own solution gives an energy estimate, which yields bounds independent of the Galerkin dimension in $L^\infty(0,T;H)$ and $L^2(0,T;V)$. Weak compactness then produces a limit, and the variational formulation identifies this limit as a weak solution. Finally, the Hilbert-triple time-[continuity theorem](/theorems/1145) gives an $H$-continuous representative and transfers the initial condition to the limit. [/proofplan]

[step:Choose Galerkin subspaces and solve the projected systems] Since $V$ is separable and densely embedded in $H$, choose a sequence $(w_k)_{k=1}^{\infty}$ in $V$ that is orthonormal in $H$ and whose linear span is dense in $V$. For each $m \in \mathbb{N}$, define the finite-dimensional subspace \begin{align*} W_m := \operatorname{span}\{w_1,\dots,w_m\} \subset V. \end{align*} Let $P_m:H \to W_m$ denote the $H$-[orthogonal projection](/theorems/437). For each $m \in \mathbb{N}$, seek a map $u_m:[0,T]\to W_m$ of the form \begin{align*} u_m(t)=\sum_{j=1}^m d_{mj}(t)w_j. \end{align*} The coefficient functions $d_{mj}:[0,T]\to \mathbb{R}$ are required to solve, for $i=1,\dots,m$, \begin{align*} (u_m'(t),w_i)_H+B_t[u_m(t),w_i]=f(t)(w_i) \end{align*} for $\mathcal{L}^1$-a.e. $t\in(0,T)$, with initial condition \begin{align*} u_m(0)=P_m u_0. \end{align*} Equivalently, the vector $d_m=(d_{m1},\dots,d_{mm}):[0,T]\to\mathbb{R}^m$ satisfies the linear system \begin{align*} d_{mi}'(t)+\sum_{j=1}^m B_t[w_j,w_i]d_{mj}(t)=f(t)(w_i). \end{align*} The coefficient functions $t\mapsto B_t[w_j,w_i]$ are measurable, and the forcing functions $t\mapsto f(t)(w_i)$ belong to $L^2(0,T;\mathbb{R}) \subset L^1(0,T;\mathbb{R})$. Hence the finite-dimensional Caratheodory existence theorem for linear systems gives an absolutely continuous solution $d_m:[0,T]\to\mathbb{R}^m$. Therefore $u_m$ is an absolutely continuous map from $[0,T]$ to $W_m$, and the projected equation holds for every $v_m\in W_m$: \begin{align*} (u_m'(t),v_m)_H+B_t[u_m(t),v_m]=f(t)(v_m) \end{align*} for $\mathcal{L}^1$-a.e. $t\in(0,T)$. [/step]

[step:Derive the uniform energy estimate]Take $v_m=u_m(t)$ in the Galerkin equation. Since $u_m:[0,T]\to W_m$ is absolutely continuous and $W_m$ is finite-dimensional, the map $t\mapsto \|u_m(t)\|_H^2$ is absolutely continuous and \begin{align*} (u_m'(t),u_m(t))_H=\frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2 \end{align*} for $\mathcal{L}^1$-a.e. $t\in(0,T)$. Thus \begin{align*} \frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2+B_t[u_m(t),u_m(t)]=f(t)(u_m(t)). \end{align*} Using coercivity and the dual norm inequality gives \begin{align*} \frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2+\alpha\|u_m(t)\|_V^2-\beta\|u_m(t)\|_H^2 \leq \|f(t)\|_{V^*}\|u_m(t)\|_V. \end{align*} [Young's inequality](/theorems/244) with parameter $\alpha$ gives \begin{align*} \|f(t)\|_{V^*}\|u_m(t)\|_V \leq \frac{\alpha}{2}\|u_m(t)\|_V^2+\frac{1}{2\alpha}\|f(t)\|_{V^*}^2. \end{align*} Therefore \begin{align*} \frac{d}{dt}\|u_m(t)\|_H^2+\alpha\|u_m(t)\|_V^2 \leq 2\beta\|u_m(t)\|_H^2+\frac{1}{\alpha}\|f(t)\|_{V^*}^2. \end{align*} By Gronwall's inequality applied to $y_m(t):=\|u_m(t)\|_H^2$, \begin{align*} \|u_m(t)\|_H^2 \leq e^{2\beta T}\left(\|P_m u_0\|_H^2+\frac{1}{\alpha}\int_0^{\mathsf T} \|f(s)\|_{V^*}^2\,d\mathcal{L}^1(s)\right). \end{align*} Since $P_m$ is an $H$-orthogonal projection, $\|P_m u_0\|_H\leq \|u_0\|_H$. Define \begin{align*} C_0:=e^{2\beta T}\left(\|u_0\|_H^2+\frac{1}{\alpha}\|f\|_{L^2(0,T;V^*)}^2\right). \end{align*} Then \begin{align*} \operatorname{ess\,sup}_{0<t<T}\|u_m(t)\|_H^2 \leq C_0. \end{align*} Integrating the differential inequality from $0$ to $T$ and using the preceding bound gives \begin{align*} \alpha\int_0^{\mathsf T} \|u_m(t)\|_V^2\,d\mathcal{L}^1(t) \leq \|u_0\|_H^2+2\beta T C_0+\frac{1}{\alpha}\|f\|_{L^2(0,T;V^*)}^2. \end{align*} Thus $(u_m)$ is bounded in $L^\infty(0,T;H)$ and in $L^2(0,T;V)$ independently of $m$.[/step]

[guided]The energy estimate is the point where the coercivity of $B_t$ is converted into compactness. We test the finite-dimensional equation with the solution itself, which is legitimate because $u_m(t)\in W_m$ for every $t$. The Galerkin identity gives \begin{align*} (u_m'(t),u_m(t))_H+B_t[u_m(t),u_m(t)]=f(t)(u_m(t)). \end{align*} Because $u_m$ is absolutely continuous as a map into the finite-dimensional space $W_m$, the squared $H$-norm is absolutely continuous and satisfies \begin{align*} (u_m'(t),u_m(t))_H=\frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2 \end{align*} for $\mathcal{L}^1$-a.e. $t$. Substituting this identity into the Galerkin equation gives \begin{align*} \frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2+B_t[u_m(t),u_m(t)]=f(t)(u_m(t)). \end{align*} Now we use the two hypotheses that control the equation. Coercivity says \begin{align*} B_t[u_m(t),u_m(t)] \geq \alpha\|u_m(t)\|_V^2-\beta\|u_m(t)\|_H^2. \end{align*} The forcing term is controlled by the definition of the $V^*$ norm: \begin{align*} |f(t)(u_m(t))| \leq \|f(t)\|_{V^*}\|u_m(t)\|_V. \end{align*} Combining these two estimates yields \begin{align*} \frac{1}{2}\frac{d}{dt}\|u_m(t)\|_H^2+\alpha\|u_m(t)\|_V^2-\beta\|u_m(t)\|_H^2 \leq \|f(t)\|_{V^*}\|u_m(t)\|_V. \end{align*} The term on the right still contains $\|u_m(t)\|_V$, so we split it using Young's inequality. With the parameter $\alpha>0$, \begin{align*} \|f(t)\|_{V^*}\|u_m(t)\|_V \leq \frac{\alpha}{2}\|u_m(t)\|_V^2+\frac{1}{2\alpha}\|f(t)\|_{V^*}^2. \end{align*} This choice is made so that half of the coercive term can absorb the $\|u_m(t)\|_V^2$ contribution from the forcing. After absorption, we obtain \begin{align*} \frac{d}{dt}\|u_m(t)\|_H^2+\alpha\|u_m(t)\|_V^2 \leq 2\beta\|u_m(t)\|_H^2+\frac{1}{\alpha}\|f(t)\|_{V^*}^2. \end{align*} Set $y_m:[0,T]\to[0,\infty)$ by $y_m(t)=\|u_m(t)\|_H^2$. Dropping the non-negative term $\alpha\|u_m(t)\|_V^2$ gives \begin{align*} y_m'(t)\leq 2\beta y_m(t)+\frac{1}{\alpha}\|f(t)\|_{V^*}^2. \end{align*} Gronwall's inequality applies because $y_m$ is absolutely continuous and $\|f\|_{V^*}^2\in L^1(0,T;\mathbb{R})$. Hence \begin{align*} \|u_m(t)\|_H^2 \leq e^{2\beta T}\left(\|P_m u_0\|_H^2+\frac{1}{\alpha}\int_0^{\mathsf T} \|f(s)\|_{V^*}^2\,d\mathcal{L}^1(s)\right). \end{align*} The projection $P_m$ is orthogonal in $H$, so $\|P_m u_0\|_H\leq \|u_0\|_H$. This gives the uniform $L^\infty(0,T;H)$ bound. Returning to the differential inequality and integrating over $(0,T)$ then gives the uniform $L^2(0,T;V)$ bound: \begin{align*} \alpha\int_0^{\mathsf T} \|u_m(t)\|_V^2\,d\mathcal{L}^1(t) \leq \|u_0\|_H^2+2\beta T C_0+\frac{1}{\alpha}\|f\|_{L^2(0,T;V^*)}^2. \end{align*} Thus the Galerkin sequence is bounded in exactly the spaces needed for weak compactness.[/guided]

[step:Extract weak limits and identify the limit operator] By weak compactness in the [Hilbert space](/page/Hilbert%20Space) $L^2(0,T;V)$ and weak-star compactness in $L^\infty(0,T;H)$, after passing to a subsequence there exists \begin{align*} u \in L^2(0,T;V)\cap L^\infty(0,T;H) \end{align*} such that \begin{align*} u_m \rightharpoonup u \quad \text{in } L^2(0,T;V) \end{align*} and \begin{align*} u_m \overset{*}{\rightharpoonup} u \quad \text{in } L^\infty(0,T;H). \end{align*} For each $m$, define $A_m:(0,T)\to V^*$ by $A_m(t):=A(t)u_m(t)$. The boundedness of $B_t$ gives \begin{align*} \|A_m(t)\|_{V^*}\leq M\|u_m(t)\|_V \end{align*} for $\mathcal{L}^1$-a.e. $t$, so $(A_m)$ is bounded in $L^2(0,T;V^*)$. Passing to a further subsequence, there exists $g\in L^2(0,T;V^*)$ such that \begin{align*} A_m \rightharpoonup g \quad \text{in } L^2(0,T;V^*). \end{align*} We show that $g(t)=A(t)u(t)$ for $\mathcal{L}^1$-a.e. $t$. Fix $k\in\mathbb{N}$ and $\varphi\in L^2(0,T;\mathbb{R})$. The map $\Lambda_{k,\varphi}:L^2(0,T;V)\to\mathbb{R}$ defined by \begin{align*} \Lambda_{k,\varphi}(z):=\int_0^{\mathsf T} \varphi(t)B_t[z(t),w_k]\,d\mathcal{L}^1(t) \end{align*} is bounded and linear. Hence [weak convergence](/page/Weak%20Convergence) of $u_m$ in $L^2(0,T;V)$ gives \begin{align*} \int_0^{\mathsf T} \varphi(t)B_t[u_m(t),w_k]\,d\mathcal{L}^1(t)\to \int_0^{\mathsf T} \varphi(t)B_t[u(t),w_k]\,d\mathcal{L}^1(t). \end{align*} On the other hand, weak convergence of $A_m$ in $L^2(0,T;V^*)$ gives \begin{align*} \int_0^{\mathsf T} \varphi(t)(A_m(t))(w_k)\,d\mathcal{L}^1(t)\to \int_0^{\mathsf T} \varphi(t)g(t)(w_k)\,d\mathcal{L}^1(t). \end{align*} Since $(A_m(t))(w_k)=B_t[u_m(t),w_k]$, we obtain \begin{align*} \int_0^{\mathsf T} \varphi(t)g(t)(w_k)\,d\mathcal{L}^1(t)=\int_0^{\mathsf T} \varphi(t)B_t[u(t),w_k]\,d\mathcal{L}^1(t). \end{align*} Because $\varphi$ and $k$ are arbitrary and the span of $(w_k)$ is dense in $V$, it follows that $g(t)=A(t)u(t)$ in $V^*$ for $\mathcal{L}^1$-a.e. $t$. [/step]

[step:Pass to the limit in the variational identity] Fix $k\in\mathbb{N}$ and let $\zeta:[0,T]\to\mathbb{R}$ be continuously differentiable with $\zeta(T)=0$. For all $m\geq k$, the Galerkin identity with test vector $w_k$ gives \begin{align*} (u_m'(t),w_k)_H+B_t[u_m(t),w_k]=f(t)(w_k) \end{align*} for $\mathcal{L}^1$-a.e. $t\in(0,T)$. Multiplying by $\zeta(t)$ and integrating by parts in time gives \begin{align*} -\int_0^{\mathsf T} (u_m(t),w_k)_H\zeta'(t)\,d\mathcal{L}^1(t)+\int_0^{\mathsf T} B_t[u_m(t),w_k]\zeta(t)\,d\mathcal{L}^1(t)=(P_m u_0,w_k)_H\zeta(0)+\int_0^{\mathsf T} f(t)(w_k)\zeta(t)\,d\mathcal{L}^1(t). \end{align*} Passing to the limit uses the weak-star convergence in $L^\infty(0,T;H)$ for the first integral, the weak convergence in $L^2(0,T;V)$ for the second integral, and $P_m u_0\to u_0$ in $H$. We obtain \begin{align*} -\int_0^{\mathsf T} (u(t),w_k)_H\zeta'(t)\,d\mathcal{L}^1(t)+\int_0^{\mathsf T} B_t[u(t),w_k]\zeta(t)\,d\mathcal{L}^1(t)=(u_0,w_k)_H\zeta(0)+\int_0^{\mathsf T} f(t)(w_k)\zeta(t)\,d\mathcal{L}^1(t). \end{align*} Since the span of $(w_k)$ is dense in $V$, the same identity holds with $w_k$ replaced by any $v\in V$. Therefore $u$ has [distributional derivative](/page/Distributional%20Derivative) \begin{align*} u'=f-A(\cdot)u \end{align*} as an element of $L^2(0,T;V^*)$. Hence \begin{align*} u'(t)+A(t)u(t)=f(t) \end{align*} in $V^*$ for $\mathcal{L}^1$-a.e. $t\in(0,T)$. [/step]

[step:Recover the initial condition and the continuous representative] We have shown that $u\in L^2(0,T;V)$ and $u'=f-A(\cdot)u\in L^2(0,T;V^*)$. By the Hilbert-triple time-continuity theorem, $u$ has a representative in $C([0,T];H)$, and for every $v\in V$ the scalar map $t\mapsto (u(t),v)_H$ is absolutely continuous with derivative $u'(t)(v)$ for $\mathcal{L}^1$-a.e. $t$. The limiting variational identity from the previous step gives, for every $v\in V$ and every $\zeta\in C^1([0,T];\mathbb{R})$ with $\zeta(T)=0$, \begin{align*} -\int_0^{\mathsf T} (u(t),v)_H\zeta'(t)\,d\mathcal{L}^1(t)+\int_0^{\mathsf T} (A(t)u(t))(v)\zeta(t)\,d\mathcal{L}^1(t)=(u_0,v)_H\zeta(0)+\int_0^{\mathsf T} f(t)(v)\zeta(t)\,d\mathcal{L}^1(t). \end{align*} Using $u'=f-A(\cdot)u$ and integrating by parts for the absolutely continuous scalar function $t\mapsto (u(t),v)_H$, the same left-hand side equals \begin{align*} (u(0),v)_H\zeta(0)+\int_0^{\mathsf T} f(t)(v)\zeta(t)\,d\mathcal{L}^1(t). \end{align*} Therefore \begin{align*} (u(0),v)_H=(u_0,v)_H \end{align*} for every $v\in V$. Since $V$ is dense in $H$, this implies $u(0)=u_0$ in $H$. Thus the continuous representative $u\in C([0,T];H)$ satisfies the evolution equation and the prescribed initial condition. [/step]