Let $\left(f_{n}\right)_{n \geq 1}$ be a sequence of continuous functions $f_{n}: \mathbb{R} \rightarrow[0,+\infty) .$ Suppose that the series $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly in an interval $(a, b)$ and $\sum_{n=1}^{\infty} f_{n}(y)<\infty$ for $y \in\{a, b\} .$ Prove that $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly on $[a, b]$
Can I deduce that $\sum_{n=1}^{\infty}f_n(a),~\sum_{n=1}^{\infty}f_n(b),$ are uniformely Cauchy by using the boundedness, continuity and the uniformity in convergence on $(a,b)$? How to apply Weierstrass M test here?
Best Answer
Your first idea is correct.
Since convergence is uniform for $y \in (a,b)$, for any $\epsilon > 0$ there exists $N(\epsilon) \in \mathbb{N}$ such that for all $m > n > N(\epsilon)$ and $y \in (a,b)$, we have
$$\tag{*}\left|\sum_{k=n+1}^m f_k(y) \right| < \epsilon$$
Pointwise convergence of the series at the endpoints also guarantees the existence of $N_a(\epsilon), N_b(\epsilon)$ such that (*) holds for all $m > n > \max(N_a(\epsilon),N_b(\epsilon))$ and $y \in \{a,b\}$.
Thus, (*) holds for all $m > n > \max(N(\epsilon),N_a(\epsilon),N_b(\epsilon))$ and $y \in [a,b]$ and the series converges uniformly on $[a,b]$ by the uniform Cauchy criterion.