Consider the sequence of functions $\{Q_n\}_{n=1}^\infty$ defined in the proof of Rudin's theorem 7.26:
$$Q_n(x)=c_n(1-x^2)^n$$
where coefficient $c_n$ is choosen so that $\int_{-1}^1 Q_n(x)dx=1$ for each $n\in\mathbb N$. I wish to show that $\{Q_n\}$ is not equicontinuous.
Here is my attempt:
Pick $\epsilon=\frac{c_1}{2}$. Let $\delta>0$. Choose $x=0$ and $y=\frac{\delta}{2}$. Since $Q_n\to0$ uniformly on $[-1,-\frac{\delta}{2}]\cup[\frac{\delta}{2},1]$, there is $N\in\mathbb N$ s.t. $|f_N(x)|<\frac{c_1}{2}$ for any $x\in[-1,-\frac{\delta}{2}]\cup[\frac{\delta}{2},1]$. In particular, $f_N(y)<\frac{c_1}{2}$. Then, $|f_N(x)-f_N(y)|=c_N-f_N(y)>c_N-\frac{c_1}{2}>\frac{c_1}{2}=\epsilon$.
Does it look good? I am not sure how to justify that $c_1<c_2<\cdots$.