I'm studying uniform convergence of harmonic function. In the course of studying, There is a part that I cannot understand. It is as follows
Suppose that $u_n(x,y)$ are $C^1$ functions on an open set $R$ for $n \ge 0$ such that $\sum_{n=0}^{\infty}u_n(x,y)$ converges uniformly to u(x,y) and $\sum_{n=0}^{\infty}\frac{\partial}{\partial x}u_n(x,y)$ converges uniformly to $v(x,y)$. Then $\frac{\partial}{\partial x}u_(x,y)=v(x,y)$
pf. It is enough to verify the theorem on a small square about an arbitrary point $(x_0,y_0)$ in R. By limiting ourselves to a square, the whole line segment from $(x_0,y)$ to $(x,y)$ will lie in R. We define function on this small square by
$$ w_n(x,y)=\sum_{k=0}^{n}u_k(x,y)=\sum_{k=0}^{n}u_k(x_0,y)+\int_{x_0}^{x}**\frac{\partial}{\partial x}**\sum_{k=0}^{n}u_k(t,y)dt$$
$$w(x,y)=u(x_0,y)+\int_{x_0}^{x}v(t,y)dt$$
since the integrands $\frac{\partial}{\partial x}\sum_{k=0}^{n}u_k(t,y)$ converge uniformly to $v(t,y)$ Corollary 8.3.2 shows that $w_n(x,y)$ converges uniformly to $w(x,y)$. But this limit is $u(x,y). Therefore, by the Fundamental Theorem of Calculas,
$$\frac{\partial}{\partial x}u(x,y)=\frac{\partial}{\partial x}w(x,y)=\frac{\partial}{\partial x}\big(u(x_0,y)+\int_{x_0}^{x}v(t,y)dt\big)(x,y)=v(x,y)$$
I have some question above proof.
Q1. Why do we suffice to "verify the theorem on a small square about an arbitrary point$(x_0,y_0)$"?
Q2. Is that correct above notation "$\int_{x_0}^{x}\frac{\partial}{\partial x}\sum_{k=0}^{n}u_k(t,y)dt$"?
I think $\frac{\partial}{\partial x}$ may be changed for $\frac{\partial}{\partial t}$ right?
I will wait your help. Thank you!
cf corol 8.3.2 