Consider a Gelfand triple $V\subseteq H$, together $f=f_1+f_2 \in L^2(0,T,H) + H^1(0,T,V^{*}) $ and a symmetric and bounded, coercive operator $A: V\rightarrow V^*$.
Consider the heat equation
$$u'+Au=f$$ $$u(0)=0$$
I know (and have proved) that under the given hypothesis, one gets $u'\in L^2(0,T,H)$.
Are you aware of a good reference for this fact?
I need to cite this result in a paper, so as to avoid putting a proof of this fact.
Sketch of proof
I will refer to this set of notes. We assume $V$ is separable, hence, take $V_1\subset V_2 \subset... \subset V$ such that $\bigcup_n V_n$ is dense in $V$. Choose a basis $v_1^n,...,v_{m_n}^n$ of $V_n$ and represent $u$ as $u_n = \sum_{j=1}^{m_n}u_jv_j$. Here, $u_j=u_j(t)$.
We use the Galerkin method and find a hopefully convergent sequence if $u_n$ satisfying $$\langle u_n'+Au_n-f,v \rangle_{V^*,V} = 0, \text{ for all } v\in V_n \text{ and a.e. in }(0,T)$$
Lemma. We can then reason as in 1 (see (57) at page 23, but I believe this is standard) to obtain
$$||u_n||_{L^2(0,T,V)}^2\leq||u_n||_{C([0,T],H)}^2+||u_n||_{L^2(0,T,V)}^2\leq c ||f||_{L^2(0,T,V^*)}^2$$
Proof of the claimed result (with $f_1=0$). Test the defition of $u_n$ with $u_n'$ and integrate:
$$\int_0^t||u_n'||_H^2+\int_0^t \langle Au_n,u_n'\rangle_{V^*,V}=\int_0^t \langle f_2,u_n'\rangle_{V^*,V}$$
Let $\alpha$ be the coercivity constant of $A$. We find
$$\int_0^t \langle Au_n,u_n'\rangle_{V^*,V} \geq \frac{\alpha}{2}||u_n(t)||_V^2$$
By integrating by parts we see
$$\int_0^t \langle f_2,u_n'\rangle_{V^*,V} = -\int_0^t \langle f_2',u_n\rangle_{V^*,V} + \langle f_2(t),u_n(t)\rangle_{V^*,V}$$
Hence Young's inequality yields, for $0<\epsilon<\alpha/2$:
$$|\int_0^t \langle f_2,u_n'\rangle_{V^*,V}| = \frac{1}{2}||u_n||^2_{L^2(0,T,V)} + \frac{1}{2}||f_2'||^2_{L^2(0,T,V^*)} + \epsilon||u_n(t)||_V^2+ \epsilon^{-1}||f_2(t)||^2_{V^*}$$
Combining everything we get $$\int_0^t||u_n'||_H^2\leq (\frac{\alpha}{2}-\epsilon)||u_n(t)||_V^2 + \int_0^t||u_n'||_H^2\leq \frac{1}{2}||u_n||^2_{L^2(0,T,V)} + \frac{1}{2}||f_2'||^2_{L^2(0,T,V^*)}$$
Using the lemma, we see that $u_n'$ weakly converges to something in $L^2(0,T,H)$, modulo subsequences. But one can show (see e.g. the lemma again, and the fact that $A$ is bounded) that the whole $u_n'$ converges to $u'$, weakly and in the $L^2(0,T,V^*)$ sense. Therefore, $u'\in L^2(0,T,H)$.