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 $(0,1)$?


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 $n \to \infty.$

  • $\begingroup$ Right, wouldn't this imply that the claim is true? $\endgroup$ – CuriousKid7 Jan 31 '17 at 21:28
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    $\begingroup$ Yes it is true. $\endgroup$ – RRL Jan 31 '17 at 21:30
  • $\begingroup$ What about $\sum x^n$ on $[a,b]$ for $0<a<b<1$? $\endgroup$ – CuriousKid7 Feb 1 '17 at 2:46
  • $\begingroup$ Yes that is uniformly convergent. By Weierstrass test since $\sum b^n$ is a convergent geometric series when $|b| < 1$ $\endgroup$ – RRL Feb 1 '17 at 3:15
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    $\begingroup$ As long as $b < 1$ $\endgroup$ – RRL Feb 1 '17 at 3:22

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