6.4.2 (a) If $\sum_{n=1}^\infty g_n$ converges uniformly, then $(g_n)$ converges uniformly to zero.

Proof: Let $\varepsilon > 0$ arbitrary. By Cauchy Criterion for Uniform Convergence of Series, there exists some $N \in \mathbb{N}$ such that $|g_{m+1}(x) + \dotsb + g_n(x)| < \varepsilon$ whenever $n > m \geq N$. Since this holds for all $m \geq N$, we can set $n = m+1$ and get $|f_n(x) |< \varepsilon$ whenever $n > N$. This proves $g_n$ uniformly converges to $0$.

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I don't understand the last two sentences,

1) Why do we get $|f_n(x)|$ by setting $n = m+1$?

2) How does $|f_n(x)| < \varepsilon$ imply $g_n$ uniformly converges to 0?

  • $\begingroup$ Seems like it is a typo, $f_n=g_n$. $\endgroup$ – Ton Jun 11 '18 at 0:06
  • $\begingroup$ @Ton that's also what I guessed but wasn't sure whether there was necessity of having another function of series. How about 2)? It's clear that $|g_n|$ converges uniformly by Weierstrass M-Test, but how do we know it uniformly converges to 0? $\endgroup$ – mathnub Jun 11 '18 at 0:19
  • $\begingroup$ @mathnub Because $|f_n(x)-0|=|f_n(x)|<\varepsilon$ for all $n>N$, so as you may recognize, it does by definition. $\endgroup$ – The Phenotype Jun 11 '18 at 0:27
  • $\begingroup$ @ThePhenotype That's so true lol (what I meant to ask was whether it was a definition, property, theorem, or else) $\endgroup$ – mathnub Jun 11 '18 at 3:56

1) This is a typo, it should be $|g_n(x)| < \varepsilon$ for all $n > N$, which does follows from the previous estimate by setting $n = m+1$.

2) We have seen that for every $\varepsilon > 0$ there exists some $N \in \mathbb{N}$ with $|g_n(x) - 0| < \varepsilon$ for all $n > N$ and all $x$. This is precisely the definition of $g_n$ converging uniformly to $0$.


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