Let $V$ be Hilbert and separable. Suppose $f \in L^2(0,T;V')$. I want to show that if $$\langle f, v \rangle=0$$ holds for all $v \in L^2(0,T;V)$, then $$\langle f(t), w(t) \rangle_{V',V} = 0\quad\text{for almost every $t \in (0,T)$}$$ holds for all $w \in L^2(0,T;V)$.
Attempted proof: Let $v \in L^2(0,T;V).$ Then $$\langle f, v \rangle = \int_0^T \langle f(t), v(t) \rangle_{V',V}.$$ Now pick $v = \varphi w$, where $\varphi \in C_c^\infty(0,T)$ and $w \in L^2(0,T;V)$. Then the above becomes $$\langle f, v \rangle = \int_0^T \varphi(t)\langle f(t), w(t) \rangle_{V',V}$$ which holds for all $w \in L^2(0,T;V)$ and for all $\varphi \in C_c^\infty(0,T)$. Then by a well-known result, we find $$\langle f(t), w(t) \rangle_{V',V}=0 \quad\text{for almost every $t \in (0,T)$}$$ and for every $w \in L^2(0,T;V)$.
Is this right? I am not sure I can pick $v=\varphi w$ like that.. what if $w$ contains a term $\frac{1}{\varphi}?$
I'm not sure if I'm missing the point of the question:
First we identify $V'$ with $V$ and $L^2(0,T,V)$ with $L^2(0,T,V')$ via the Riesz Representation. Suppose that $\langle f, v \rangle = 0$ for all $v \in L^2(0,T,V)$. Then in particular $$\int_0^T \! \|f(t)\|^2 \, dt = \langle f, f \rangle = 0$$ Hence $\|f(t)\| = 0$ and thus $f(t) = 0$ almost everywhere. Thus $\langle f(t), w(t) \rangle = 0$ almost everywhere for any $w \in L^2(0,T,V)$.