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 $t\in (0,1)$?
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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 $x=1/2$).
answered May 16, 2023 at 15:30