Let $H$ be a Hilbert space with basis $b_i$. For all $t$, let $f(t;\cdot,\cdot)$ be an inner product on $H$. For each $j$, is $$\int_0^T \sum_{i=1}^\infty f(t,b_i,b_j)x_j(t)=0$$ uniquely solvable for $x_j$, given that $$\sum_{i=1}^\infty f(t,b_i,b_j)x_j(t)=0$$ is uniquely solvable for each $t$? (This second displayed equation is not given to hold by the way).
Basically I want to show that $x_j=0$ for all $j$.
We have that $t \mapsto f(t;\cdot,\cdot)$ is integrable.