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 ?

  • $\begingroup$ is $K$ different from $\Bbb{K}$? $\endgroup$
    – graydad
    Jan 27, 2015 at 18:14
  • $\begingroup$ yea ofcourse, $K$ is a compact space, $\mathbb{K}$ is a field of complex or real numbers. $\endgroup$
    – J.E.M.S
    Jan 27, 2015 at 18:16
  • $\begingroup$ 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. $\endgroup$
    – Pp..
    Jan 27, 2015 at 20:26
  • $\begingroup$ 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 ... $\endgroup$
    – J.E.M.S
    Jan 27, 2015 at 20:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – Pp..
    Jan 27, 2015 at 21:07

3 Answers 3


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

  • $\begingroup$ 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. $\endgroup$
    – Pp..
    Jan 30, 2015 at 21:19
  • $\begingroup$ @ 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 $C(K)$. $\endgroup$
    – Teri
    Jan 31, 2015 at 9:00
  • $\begingroup$ @Teri very nice solution, many thanks. $\endgroup$
    – J.E.M.S
    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


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 $$\sum_{n=1}^{\infty}|a_n|=|\mu|(K)<\infty.$$

  • $\begingroup$ 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. $\endgroup$ Jan 30, 2015 at 21:22
  • $\begingroup$ @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. $\endgroup$
    – 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:)

  • $\begingroup$ To choose $U_{31},U_{32}$ you may need to shrink $U_1,U_2$ and therefore change $f_1,f_2$. $\endgroup$
    – Pp..
    Jan 27, 2015 at 20:01
  • $\begingroup$ @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! $\endgroup$
    – Haha
    Jan 27, 2015 at 20:03
  • $\begingroup$ 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 $K$. $\endgroup$
    – Pp..
    Jan 27, 2015 at 20:06
  • $\begingroup$ @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 ?) $\endgroup$
    – J.E.M.S
    Jan 27, 2015 at 20:12
  • $\begingroup$ @Pp.., please take a look, I corrected the regions... $\endgroup$
    – Haha
    Jan 27, 2015 at 20:22

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