Let $f$ be continuous on $[-1,1]$, Show that $g_n \to f$ uniformly on $[-1,1]$ where
$$ g_n(x) = \int_{-1}^{1}f(y)p_n(y - x)dy $$
and
$$ p_n(x) = \frac{(1-x^2)^n}{\int_{-1}^{1}(1-x^2)^ndx} $$
I was trying to show that
$$ \lim_{n\to\infty}\sup_{x \in [-1,1]}|g_n - f| = 0 $$
I permute the sums and integral and use the binomial theorem. Also, I use the fact that $f$ is bounded (continuous on a compact set). After all of that, I could only get an upper bound for the sup. But I can not show that the bound goes to $0$ as $n$ goes to $\infty$. There is a simpler approach? Thanks.
I think the idea here is that $p_n\geq 0$ is a very good approximation to the delta function meaning: for all $\delta>0$, $\lim_{n\to\infty}\int_{|y|>\delta}p_n(y)dy=0$, $\int p_n\equiv 1$ and $\sup_n\|p_n\|_{L^1}=1<+\infty$.
Now given $\varepsilon>0$, choose $\delta>0$ so that $|x-y|\leq\delta\implies|f(x)-f(y)|\leq\varepsilon$ by uniform continuity of $f$. And choose $N$ so that $n>N\implies \int_{|y|>\delta}p_n(y)dy\leq\varepsilon$. Then, \begin{align} |g_n(x)-f(x)| &\leq \int_{-1}^1 |f(y-x)-f(x)|p_n(y)dy\\ &=\int_{0\leq|y|\leq\delta} |f(y-x)-f(x)|p_n(y)dy+\int_{1\geq|y|>\delta} |f(y-x)-f(x)|p_n(y)dy\\ &\leq\varepsilon\cdot\|p_n\|_{L^1}+2\|f\|_{\infty}\cdot\varepsilon\quad\text{(Holder's inequality)}\\ &=\varepsilon\cdot(1+2\|f\|_\infty). \end{align}
Since $\varepsilon>0$ was arbitrary and the upper bound is independent of $x$, we are done.