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I dont have totally clear the meaning of the expression

$$\sup_{n\in\Bbb N}\left\|\sum_{k=0}^ng_k\right\|_\infty<\infty$$

for a series of functions $g_k$ valued in $\Bbb R$ or $\Bbb C$. In particular I dont know if the supremum over the naturals have the infinity included as a "limit point" or so. The difference is very important:

  • if infinity is a "limit point" of $\Bbb N$ then the above expression doesnt necessarily imply that the series eventually decreases or remain constant.

  • if infinity is not a limit point of $\Bbb N$ then the above expression imply that the series eventually decreases or stay constant.

Unfortunately the context where I get this (an exercise) dont show clearly what is the exact meaning. I would like to assume the second, what would simplify the exercise dramatically, but Im not sure. Some help will be appreciated.

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    Doesn't this mean the sum is bounded? – 2017-01-27
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    In what space do the $g_k$ lie, and what is the norm $\|\cdot\|_{\infty}$ ? – 2017-01-27
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    @SimplyBeautifulArt of course, in any case the series converges and the partial sums are bounded. But I dont see clearly if the convergence imply that the series eventually decrases or remain constant. – 2017-01-27
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    @MPW I will update, the space is $\Bbb C$ or $\Bbb R$. – 2017-01-27

2 Answers 2

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I'm not quite sure what you mean by including infinity as a limit point of $\mathbb{N}$ but I will offer you my interpretation of what the expression is telling you.

It seems to me that $\{g_k\}$ is a sequence of functions, and essentially here we've defined another sequence of functions, say $f_n$, by $$f_n(x)=\sum_{k=1}^{n}g_k(x).$$ So the expression merely says that $\sup_{n\in\mathbb{N}}||f_n||_\infty <\infty$. That is, $f_n$ is uniformly bounded by some finite $M<\infty$. Then, in particular, for all $x\in\mathbb{R}$ or $\mathbb{C}$ or whatever the underlying domain is, and for all $n$ we must have $$|f_n(x)|=|g_1(x)+...+g_n(x)|\leq M.$$ As for the individual $g_k$, it follows that $|g_k(x)|=|f_{k}(x)-f_{k-1}(x)|\leq |f_{k}(x)|+|f_{k-1}(x)|\leq 2M<\infty$. With all that said, I don't think you can say much about the increasing/decreasing nature of $f_n$.

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This is just the supremum of a countable set of real numbers, right? Written differently, it is $$\sup\left\{ \|S_1\|_{\infty}, \|S_2\|_{\infty}, \|S_3\|_{\infty}, \ldots \right\}$$ where $S_n$ is the $n^{\textrm{th}}$ partial sum: $S_n=\sum\limits_{k=0}^n g_k$. And the statement is just that this supremumum is finite.

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    Ok, then it is the second meaning, as I thought, right? Exists some $N\in\Bbb N$ for the supremum. Hence the series decreases or stay constant for $n\ge N$. Right? – 2017-01-27
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    I don't think so. I think you could have a strictly increasing but bounded sequence, can't you? That is, you could have $$\|S_1\|<\|S_2\|<\|S_3\|<\cdots<\|S_n\|<\|S_{n+1}\|<\cdots $\|S_n\| =1-\frac1{n+1}$. – 2017-01-27
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    Oh! I see... thank you very much. I was not seeing that, despite being countable, the set of $\|S_n\|$ can have limit points because they are real values... – 2017-01-27
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    I imagine one could also come up with a sequence of functions such that $ S_i $ has the same $ \infty $-norm for all $ i $, but $ S_i \neq S_j $ for all $ i\neq j $. Maybe $ g_0(t) = \sin(t) $, $ g_k(t) = \sin(t+k) - \sin(t+(k-1)) $? – 2017-01-27