 Solution

Suppose for contradiction that some $q\in {\mathcal{P}}_n$ satisfiesAt $t_i$,Because $|f(t_i)-q(t_i)|<a_i$, the values $p(t_i)-q(t_i)$ have strictly alternating signs. The intermediate value theorem therefore gives at least one root of $p-q$ in each of the $n+1$ intervals $(t_i,t_{i+1})$. A nonzero polynomial of degree at most $n$ cannot have $n+1$ distinct roots. If $p-q$ were identically zero, its error at $t_i$ would be $a_i\ge a_{\ast }$, also a contradiction. Hence