Rudin's bounded rational spikes are not equicontinuous and have no uniformly convergent subsequence

The denominator is positive: if its two nonnegative summands vanished, $x=0$ and $1-nx=0$ would both hold, which is impossible. Each $f_n$ is therefore continuous, $0\le f_n\le 1$, $f_n(0)=0$, and $f_n(1\mathrm{/}n)=1$.

For every fixed $x>0$, $f_n(x)\to 0$, while $f_n(0)=0$. Thus any uniformly convergent subsequence would have the zero function as its pointwise limit, consistently with [L1].

Take $\epsilon =1\mathrm{/}2$. For every $\delta >0$ there is an integer $n\ge 1$ with $1\mathrm{/}n<\delta$, but $\mathrm{\mid }{f}_n(1\mathrm{/}n)-f_n(0)\mathrm{\mid }=1$. Thus no common radius works at $0$, so the family is not equicontinuous.

For every subsequence $(f_{n_j})$, the indices $n_j$ tend to infinity, so step 1.2 still forces any uniform limit to be zero. But $\parallel f_{n_j}{\parallel }_{\mathrm{\infty }}=f_{n_j}(1\mathrm{/}{n}_j)=1$ for every $j$, contradicting uniform convergence to zero. No subsequence is uniformly convergent.

This bounded family is neither equicontinuous nor uniformly sequentially compact.