Let $B: C^{\infty}([0,1]^3)$ satisfy
$$B(t,t,x)=0 \quad \text{for all $t,x \in [0,1]$.}$$
Let $f \in C^{\infty}([0,1]^2)$ satisfy the following integral equation:
$$ \int\_0^1 f(t,x)\,dx + \int\_0^t\left(\int\_0^1 f(s,x)\,B(t,s,x)\,dx\right)\,ds =0, \quad \forall\, t\in (0,1).$$
Does it follow that
$$ \int\_0^1 f(t,x)\,dx=0$$
for all $t\in (0,1)$?



On an integral equation
The answer is no. A counterexample is
$$
f(t,x) = x - \frac{1}{2} -\frac{1}{24} t^2
$$
$$
B(t,s,x) = \left( x-\frac{1}{2} \right) (t-s)
$$
(Method: I obtained this by expanding $f$ and $B$ into power series in the arguments $t$ and $s$, considering the integral equation order by order in $t$, fiddling a bit with how few nonzero expansion coefficients I could get away with, and utilizing symmetry/antisymmetry around $x=1/2$).


