= Solution
Composition with $t\mapsto t^{1+\alpha_n}$ preserves continuity, so $A_n$ maps the <space of continuous functions on a compact space> $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+\alpha_n}$ maps $[0,1]$ onto itself,
$$
\|A_nx\|_\infty=\|x\|_\infty,\qquad\|A_n\|=1.
$$
For $0<t\leq1$, the <mean value theorem> applied to $r\mapsto t^{1+r}$ gives
$$
0\leq t-t^{1+\alpha_n}\leq\alpha_n\,t|\log t|\leq\frac{\alpha_n}{e}.
$$
The same bound holds at $t=0$. Each $x\in C([0,1])$ has <uniform continuity> by the <Heine-Cantor theorem>. With its <modulus of continuity> $\omega_x(\delta)=\sup_{|s-t|\leq\delta}|x(s)-x(t)|$, we obtain
$$
\boxed{\|A_nx-x\|_\infty\leq\omega_x(\alpha_n/e)\longrightarrow0.}
$$
Alternatively, the bound proves convergence on every <polynomial>; the <Weierstrass approximation theorem> makes these a <dense subset> of $C([0,1])$, and the <Banach-Steinhaus convergence theorem> with $\|A_n\|=1$ gives the same conclusion.
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