# Series convergence and compact space

Let $K$ be a compact topological Hausdorff space. $\{x_n\}_1^\infty \subset K$ such that $x_i \not= x_j, i \not=j$ and $\{a_i\}_1^\infty \subset \mathbb{K}$. Show the folowing are equivalent:

1. for all $f \in C(K)$ the series $\sum_{n=1}^\infty a_nf(x_n)$ is convergent
2. $\sum_{n=1}^\infty |a_n|<\infty$

Implication from $2$ to $1$ is ofcourse easy but what about the second one ?

• is $K$ different from $\Bbb{K}$? Jan 27, 2015 at 18:14
• yea ofcourse, $K$ is a compact space, $\mathbb{K}$ is a field of complex or real numbers. Jan 27, 2015 at 18:16
• If we knew the linear functional $f\mapsto \sum a_nf(x_n)$ were continuous then we get $\sum |a_n|<\infty$ using the Riesz-Markov-Kakutani theorem.
– Pp..
Jan 27, 2015 at 20:26
• Hmmm I am not schure (but I don't know) wheather if the functional is well defined (1 tells that) than it is automaticaly continious. But it's some start ... Jan 27, 2015 at 20:39
• If the sequence $\{x_n\}$ had a finite number of accumulation points then we can prove it in this way. The idea is that if $\sum |a_n|=\infty$ then there is $b_n\to0^+$ such that still $\sum|a_n|b_k=\infty$. Then we can put $f(x_n)=\text{sgn}(a_n)b_n$, when $x_n$ is not an accumulation point of $\{x_n\}$ and $f(x_n)=0$ otherwise, then extend $f$ to a continuous function on $K$ using Tieztze extension theorem and we are done.
– Pp..
Jan 27, 2015 at 21:07

Let us define the functionals $\lambda_N\in C(K)^*=M(K)$ by $$\lambda_N=\sum_{n=1}^N a_n\delta_{x_n}.$$ Your first assumption implies, through the uniform boundedness principle that $$\sup_N\|\lambda_N\|_{M(K)}<\infty.$$

Now recall that $$\sup\{|\lambda_N(f)|: \|f\|_{C(K)}\le 1\}= \|\lambda_N\|_{M(K)}\le \sum_{n=1}^N|a_n|.$$ Consider the function $f$ defined on the closed set $X_N=\{x_1,\ldots, x_N\}$ by $$f(x_j)=\left\{\begin{array}{rr} \frac{|a_j|}{a_j} & a_j\ne 0 \\ 0 & a_j=0.\end{array}\right.$$

By the Tietze extension theorem we may extend $f$ to a continuous function $\tilde f$ on the whole of $K$, preserving the supremum norm, which we note to be $\|f\|_\infty \le 1$. This yields $$\|\lambda_N\|_{M(K)}\ge |\lambda_N(\tilde f)|=\sum_{n=1}^N|a_n|.$$

The condition $$\sup_N \|\lambda_N\|_{M(K)}<\infty$$ now exactly means $$\sum_{n=1}^\infty |a_n|<\infty.$$

• Uniform boundedness! That shows how long it has been since I do functional analysis. Nice. In a while I should be able to vote up again, and some more later to award the bounty.
– Pp..
Jan 30, 2015 at 21:19
• @ Pp.. Indeed, uniform boundedness is the key, and by the way you can use it to show that the limit $\lambda = \lim_{N\to\infty} \lambda_N$ does define a bounded functional in $C(K)$.
– Teri
Jan 31, 2015 at 9:00
• @Teri very nice solution, many thanks. Jan 31, 2015 at 21:06

Scroll down (or maybe soon will be up). There is a solution and is not this one.

This is as far as I have got it, a tiny bit more than my comment above.

Restriction of the problem: Assume that the set $S\subset K$, where $\{x_n\}$ accumulates, contains only finitely many elements of $X:=\{x_n\}$.

Suppose, in order to get a contradiction, that

$$\sum_{n=1}^{\infty}|a_n|=\infty$$

Then there is a sequence $b_n>0$ such that $b_n\to 0$ and $$\sum_{n=1}^{\infty}|a_n|b_n=\infty$$

Let us define $$f(x_n):=\begin{cases}\text{sgn}(a_n)b_n&\text{ for }x_n\notin S\\0&\text{ for }x_n\in S\end{cases}$$ We can extend $f$ to a continuous function by putting $f(x)=0$ for $x\in S$ and to $K$ by using Tietze extension theorem.

Then $$\sum_{n=1}^{\infty}a_nf(x_n)=\sum_{n\in\mathbb{N}:\ x_n\notin S}|a_n|b_n=\sum_{n=1}^{\infty}|a_n|b_n-\text{(a finite number)}=\infty$$

Another case: Assume that $\delta: C(K)\to \mathbb{K}$, defined by $f\stackrel{\delta}{\mapsto} \sum_{n=1}^{\infty}a_nf(x_n)$ is a continuous linear functional.

Then by Riesz-Markov-Kakutani there is a Radon measure $\mu$ on $K$, with finite total variation $|\mu|(K)<\infty$ such that $$\delta(f)=\int_K f d\mu$$

We see that $\mu=\sum_{n=1}^{\infty}a_n\delta_{x_n}$, where $\delta_{x_n}$ is the Dirac measure supported at $x_n$. Then $$\sum_{n=1}^{\infty}|a_n|=|\mu|(K)<\infty.$$

• I suspect one can show that $\delta$ as defined in your second case is always continuous under the given hypotheses. I will think about how to do this, but just on principle, one shouldn't be able to construct a discontinuous linear functional on a Banach space by such a simple formula. Jan 30, 2015 at 21:22
• @NateEldredge See Teri's answer below. I didn't have Uniform Boundedness in RAM. Probably should have think it better but it is also good to read solutions.
– Pp..
Jan 30, 2015 at 21:23

1 to 2:

If $\mathbb K=\mathbb R$.(Same thought for $\mathbb K=\mathbb C$) We are going to build a continuous function that does the job. What we want to achieve is $f(x_n)=sign (a_n)$ and then $a_n\cdot sign(a_n)=|a_n|$. We suppose that $K$ is a connected space otherwise you can do what we will do in the connected components of $K$.

We proceed by induction.

Let $U_1$ be a region of $x_1$ and $U_2$ be a region of $x_2$ such that $U_1\cap U_2= \emptyset$.

Now let $U_{31},U_{32}$ be regions of $x_3$ and $U_{11},U_{22}$ b e regions of $x_1,x_2$ respectivly such that $U_{31}\cap U_{11}=\emptyset$ and $U_{32}\cap U_{22}=\emptyset$. We take $U_3=U_{31}\cap U_{32}$ as a region of $x_3$ and $\widetilde {U_1}=U_1\cap U_{11}$ and $\widetilde {U_2}=U_2\cap U_{22}$ the new regions of $x_1,x_2$.

We continue By finding $\widetilde{U_3}$ with which we will have now problem because $\widetilde{U_3}\subset U_3$.

Let $f_1:K\to \mathbb K$ such that $f_1(x)=sign(a_1)$ if $x\in \widetilde{U_1}$ and $0$ otherwise.

Let $f_2:K\to \mathbb K$ such that $f_2(x)=sign(a_2)$ if $x\in \widetilde {U_2}$ and $0$ otherwise. and

we let $f_3:K\to \mathbb K$ such that $f_3(x)=sign(a_3)$ if $x\in \widetilde {U_3}$ and $0$ otherwise.

and so on...

Now if we let $\widetilde{f_i}$ to be the "smoothness" of $f_i$, meaning to "glue" the jump from $1$(or $-1$) to $0$, we can see that $\widetilde{f_i}$ are continuous.

Let $f=\cup \widetilde{f_i}$ and because $K$ is compact $f$ is continuous with the property $f(x_n)=sign(a_n)$.

A quick thought of mine.

Try to do the gluing if you like:)

• To choose $U_{31},U_{32}$ you may need to shrink $U_1,U_2$ and therefore change $f_1,f_2$.
– Pp..
Jan 27, 2015 at 20:01
• @Pp.., yes . you are correct. I just need to find these regions for given $n4 first and then set the functions. Will do now. Thanks! – Haha Jan 27, 2015 at 20:03 • There may not be a continuous function such that$f(x_n)=\text{sgn}(a_n)$. The sequence$\{x_n\}$may have accumulation points in many places, even in all of$K\$.
– Pp..
Jan 27, 2015 at 20:06
• @Mitsos my main problem (because attempt was symilar) was actually to prove that we can find such continuous functions (I don't know how do you "smooth" them ?) Jan 27, 2015 at 20:12
• @Pp.., please take a look, I corrected the regions...
– Haha
Jan 27, 2015 at 20:22