If $\sum g_n$ converges uniformly , then does $(g_n)$ converge uniformly to $0$?
I think that it does basically from the fact that the convergence of numerical series implies that the numerical sequence of terms goes to $0$. But I feel I may be missing something.
Is the claim true?
UPDATE: Would $\sum x^n$ be a counterexample on some compact subset of $(0,1)$?
Best Answer
Note that
$$g_n(x) = \left(\sum_{k=1}^{n} g_k(x) - g(x)\right) - \left(\sum_{k=1}^{n-1} g_k(x) - g(x) \right)$$
where both series converge uniformly to the same function $g$ on some set $D$.
Using the triangle inequality we have
$$|g_n(x)| \leqslant \left|\sum_{k=1}^{n} g_k(x) - g(x)\right| + \left|\sum_{k=1}^{n-1} g_k(x) - g(x) \right|.$$
Given uniform convergence of the series, for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n > N$and for all $x \in D$, each term on the right-hand side is smaller than $\epsilon/2$.
Therefore, for all $n > N$ and for all $x \in D$ we have $|g_n(x) - 0| < \epsilon$ and $g_n \to 0$ uniformly.
The case of $\sum x^n$ for $x$ in some compact subset of $(0,1)$ is not a counterexample. If $D \subset (0,1)$ is compact, then there exists $b < 1$ such that $|x| \leqslant b$ and $|x^n| \leqslant b^n$ for all $x \in D$. Consequently $\sum b^n$ is a convergent geometric series and $\sum x^n$ converges uniformly by the Weirstrass test. Furthermore, $b^n \to 0$ which implies $x^n \to 0$ uniformly as $n \to \infty.$