How do we justify differentiating under the integral for Poisson Integral Define $f(x,y)=(f*P_y)(x)$, where $P_y(x)=\frac{1}{\pi}\frac{y}{y^2+x^2}$, $y>0$ is the Poisson kernel. ($f$ is in $L^p$, for some $1\leq p\leq\infty$.)
In other words, $f(x,y)=\frac{1}{\pi}\int_{-\infty}^\infty f(t)\frac{y}{y^2+(x-t)^2}\,dt$.
I would want to prove $\frac{\partial}{\partial x}f(x,y)=\int_{-\infty}^\infty f(t)\frac{\partial}{\partial x}P_y(x-t)\,dt$, i.e. differentiate under the integral sign.
How do we justify the differentiation under integral sign? I know it has something to do with Lebesgue's Dominated Convergence Theorem, but I am having a hard time finding something to dominate the partial derivative.
My best try was to get $\begin{align*}
\sup_{x\in\mathbb{R}}|\frac{\partial}{\partial x}P_y(x-t)|\leq C/y^2
\end{align*}$ for some constant $C$ (using calculus computations). However, it seems that this is not good for satisfying Lebesgue's DCT.
Thanks for any help.
 A: One works locally, and finds dominating functions for $x$ and $y$ in specific ranges. We start by looking at the integrand
$$h(t,x,y) = f(t)\cdot \frac{y}{y^2 + (x-t)^2}$$
and its partial derivative
$$\frac{\partial h}{\partial x}(t,x,y) = f(t) \cdot \frac{1}{y^2 + (x-t)^2}\cdot \frac{2y(t-x)}{y^2+(x-t)^2}.$$
The last factor has modulus bounded by $1$ ($2\lvert ab\rvert \leqslant a^2 + b^2$), so we obtain the inequality
$$\biggl\lvert \frac{\partial h}{\partial x}(t,x,y)\biggr\rvert \leqslant \frac{\lvert f(t)\rvert}{y^2 + (x-t)^2}.$$
Now we fix arbitrary $\delta > 0$ and $R > 0$, and note that for $y \geqslant \delta$ and $\lvert x\rvert \leqslant R$, we have
$$\biggl\lvert \frac{\partial h}{\partial x}(t,x,y)\biggr\rvert \leqslant \frac{\lvert f(t)\rvert}{\delta^2 + (\max \{\lvert t\rvert - R, 0\})^2},$$
and the latter function belongs to $L^1(\mathbb{R})$ since $\frac{1}{\delta^2 + (\max \{\lvert t\rvert - R, 0\})^2} \in L^1(\mathbb{R}) \cap L^\infty(\mathbb{R})$.
Since the regions $\{(x,y) : y \geqslant \delta \land \lvert x\rvert \leqslant R\}$ exhaust the upper half-plane, the Poisson integral can be differentiated (with respect to $x$, but the argument for $\partial/\partial y$ is similar) under the integral on the whole (open) upper half-plane.
