Series of functions: convergence interval type vs. monotony of the function and calculation of $\mathrm{sup}_{x \in [r,\infty)}|f_n(x)|$

Question

I have two doubts regarding the normal convergence of a function series.

Consider a real functions series $f_n(x): A \subset \mathbb{R} \to \mathbb{R}$ of wich I want to study the normal convergence. $$\sum_{n \geq 0} f_n(x)$$

Suppose that I find that the series $$\sum_{n \geq 0} |f_n(x)|$$ converges pointwise in a open interval $I=(a,+ \infty)$.

Since normal convergence is stronger than pointwise convergence of $\sum_{n \geq 0} |f_n(x)|$, only $x \in I$ should be considered. Besides that, proving that series converges in a open interval (as $I$ is) would imply that it converges normally also in the corresponding closed interval, that is $I'=[a,\infty)$, but this would also imply that the series $\sum_{n \geq 0} |f_n(x)|$ converges pointwise in $x=a$ which is not true. Therefore I will look in "smaller" intervals inside $I$, so for $x \in J_r=[r,\infty)$ with $r>a$.

The normal convergence requires to find some constants $M_n$ such that $$\mathrm{sup}_{x \in [r,\infty)} |f_n(x)|<M_n \,\,\,\, \mathrm{and} \,\,\,\, \sum_{n \geq 0} M_n \,\,\, \mathrm{converges}$$

That's the situation, now I will explain my doubts.

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I noticed some link between the type of the convergence interval $I$ (pointwise convergence of $\sum_{n \geq 0} f_n(x)$, in general) and the monotony of the function $f_n(x)$, that is

1. If the function $f_n(x)$ is monotonically increasing in $I$, then $I$ is of the type $I=(-\infty,a)$, $a \in \mathbb{R}$
2. If the function $f_n(x)$ is monotonically decreasing in $I$, then $I$ is of the type $I=(a,+\infty)$, $a \in \mathbb{R}$
3. If the function $f_n(x)$ is neither increasing or decreasing (for istance it has a maximum in $I$) then $I$ is of the type $(a,b)$, $a,b \in \mathbb{R}$

I'm totally not sure about this statement, is there any link between the monotony of $f_n(x)$ and the type of interval in which $\sum_{n \geq 0} f_n(x)$ converges (in the general case, pointwise)?

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My second doubt is linked to the previous one. Suppose for istance to be in case 2. (as the situation described at the beginning). If I want to prove the normal convergence, I must consider $\mathrm{sup}_{x \in [r,\infty)}|f_n(x)|$. Which of the following two is correct?

• $\mathrm{sup}_{x \in [r,\infty)}|f_n(x)|= |f_n(r)|$
• $\mathrm{sup}_{x \in [r,\infty)} |f_n(x)| \leq |f_n(r)|$

The second one includes the first one, but the two are not the same thing. Putting a $=$ means to take $|f_n(r)|$ as the smallest of the numbers such that $|f_n(x)| \leq w \,\,\,\, \forall x \in [r,\infty)$, but is this always true? Would it be a good idea to put a $\leq$ sign, is such way to be sure that the relation holds true?

Answer 1

1) In general, if $f_n : D \to \Bbb R$ and $|f_n| \le M_n$ and $\sum M_n < \infty$, then $\sum f_n$ converges normally (and therefore pointwisely) on $D$, regardless of whether the functions are monotonic or not. Take as examples the functions $\frac {\arctan nx} {n^2}$ or $\frac {\sin nx} {2^n}$ defined on $\Bbb R$. This answers your question in the negative.

2) Since your statements at 1) are false, there isn't much to be said here. In any case, if $f :[a, \infty) \to \Bbb R$ is decreasing, then indeed $\sup \limits _{x \in [a, \infty)} f(x) = f(a)$. Notice, though, that adding a modulus completely changes things: $\sup \limits _{x \in [a, \infty)} |f(x)|$ need no longer be equal to $|f(a)|$, as shown by the example $- \Bbb e ^x : [0, \infty) \to \Bbb R$ for which $\sup (- \Bbb e ^x) = - \Bbb e ^0 = -1$, but $\sup |- \Bbb e ^x| = \sup \Bbb e ^x = \infty \ne 1$.