Can Lebesgue Dominated Convergence always be used?

Question

Suppose I want to find the derivative $$\frac{d}{dx}\int f(x,y) dy.$$ I want to know under what condition it would be equal to $$\int \frac{d}{dx}f(x,y) dy.$$ Of course, if I can find a suitable dominating function $H$, then I can use LDCT. In other words, we know that $$\text{Existence of a dominating function } H \implies \left(\frac{d}{dx}\int f(x,y) dy=\int \frac{d}{dx}f(x,y) dy\right).$$ But is this actually an if and only if? In other words, does interchangeability of the derivative and integral necessarily imply the existence of a dominating function $H$ which I can use to apply LDCT, i.e., $$\left(\frac{d}{dx}\int f(x,y) dy=\int \frac{d}{dx}f(x,y) dy\right) \implies \text{Existence of a dominating function } H.$$

Answer 1

To make this more well-defined, let's say $f$ is a function on $[-1,1] \times [-1,1]$ which is differentiable with respect to $x$ at $x=0$ for all $y$, and $f(x, \cdot) \in L^1[-1,1]$ for all $x$, and $\dfrac{\partial f}{\partial x} (0, \cdot) \in L^1[-1,1]$. Thus it makes sense to ask whether $$ \left.\dfrac{d}{d x} \int_{-1}^1 f(x,y)\; dy\right|_{x=0} = \int_{-1}^1 \dfrac{\partial f}{\partial x}(0,y)\; dy \tag{1}$$ To use Dominated Convergence, you would want some $\epsilon > 0$ and $g \in L^1$ such that $$ \left| \dfrac{f(x,y) - f(0,y)}{x}\right| \le g(y)\ \text{for}\ 0 < |x|<\epsilon \tag{2}$$

To find examples where (1) is true but not (2), you might rely on symmetry to make (1) work. For example, this will be true if $f(x,y)$ is an odd function of $y$ for every $x$. So e.g. try

$$ f(x,y) = \cases{\dfrac{x^2 y}{x^4 + y^4} & $y \ne 0$\cr 0 & $y = 0$\cr}$$ noting that $f(y,y)/y = 1/(2y^2)$ for $y \ne 0$, so that (2) fails, but both sides of (1) are $0$, indeed $$\dfrac{d}{dx} \int_{-1}^1 f(x,y)\; dy = \int_{-1}^1 \dfrac{\partial f}{\partial x}(x,y)\; dy = 0\ \text{for all}\ x$$