2
$\begingroup$

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 $(0,1)$?

$\endgroup$
4
$\begingroup$

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 $n \to \infty.$

$\endgroup$
  • $\begingroup$ Right, wouldn't this imply that the claim is true? $\endgroup$ – CuriousKid7 Jan 31 '17 at 21:28
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ As long as $b < 1$ $\endgroup$ – RRL Feb 1 '17 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.