Working on a preparing for a preliminary exam, came across the following question.
For $x=(x_1,x_2)\in\ \mathbb{R}^2$, let $\vert x\vert=(x_1^2+x_2^2)^{1/2}$. Let $D=\{x\in\mathbb{R}; \vert x\vert<1\}$, and let $f:\bar{D}\to\mathbb{R}$ be continuous on $\bar{D}$. Prove that
$$\lim_{n\to\infty}\iint_D(n+2)\vert x\vert^nf(x)\,dA=\int_0^{2\pi}f(\cos(\theta),\sin(\theta)) \, d\theta.$$
Since the domain is a circle, I used the polar transform to end up with
\begin{align} & \iint_D(n+2)\vert x\vert^n f(x)\,dA \\[8pt] = {} & \int_0^{2\pi}\int_0^1 (n+2) r^n f(r\cos(\theta), r\sin(\theta)) \cdot r\,dr\,d\theta \\ = {} & \int_0^{2\pi}\int_0^1(n+2)r^{n+1}f(r\cos(\theta),r\sin(\theta)) \, dr\,d\theta \end{align}
Since the form $(n+2)r^{n+1}$ seems like it is begging to be integrated, I began to use integration by parts, but then realized there are no differentiability assumptions about $f$. I also thought about uniform limit theorems with $f_n(r,\theta)=(n+2)r^{n+1}f(r\cos(\theta),r\sin(\theta))$, but the pointwise limit isn't continuous.
A step in the right direction would be appreciated.
You should avoid integration by parts. Let $$ \Delta(r,\theta) = f(\cos \theta,\sin \theta) - f(r\cos\theta,r\sin\theta).$$
Use compactness to show that for any $\epsilon>0$ there is $r_\epsilon<1$ so that $|\Delta(r,\theta)| <\epsilon$ for every $r_\epsilon<r\leq 1$ and uniformly in $\theta$. Use this to estimate $\int_0^1 \int_0^{2\pi} (n+2)r^{n+1}\Delta(r,\theta) d\theta\; dr $ as $n$ goes to infinity.