Does $\partial_b u(\cdot,b) \in L^2(A)$ for fixed $b$ imply that $u(\cdot,b) \in L^2(A)$?

Question

Let $u$ be defined on $A \times B$ where $A$ and $B$ are two bounded domains, and write $u=u(a,b)$.

Suppose that the weak derivative $\partial_b u(\cdot,b) \in L^2(A)$ for fixed $b$. Does this imply that $u(\cdot,b) \in L^2(A)$?

Answer 1

No.

Here is a hint on how to construct a counterexample, put $$g(x,y)= \sin(y\,h(x)),$$

then $g'_y(x,y)=h(x)\cos(y\,h(x))$.