Absolutely convergent series over all functionals

Let $X$ be a normed vector space. Let $(x_n)_{n\ge 1}$ be a sequence in $X$, such that for any $f\in X^*$
$$\sum_{n=1}^{\infty}|f(x_n)|<\infty.$$
Then there exists a constant $C$ such that
$$\sum_{n=1}^{\infty}|f(x_n)|\le C\|f\|, \ \forall f\in X^*$$.

$\textbf{Proof}.$ Consider the operator $T:X^*\to l_1$, defined by
$$T(f)=(f(x_1),f(x_2),\ldots).$$
Clearly $T$ is well-defined linear operator. Moreover $T$ is bounded. We shall check that using the Closed-graph theorem ($X^*$ is Banach, even if $X$ is not). For this purpose take a sequence $(f_k)_{k\ge 1}\subset X^*$ converging to some $f$, and let $Tf_n\rightarrow p$ for some $p\in l_1$. Our aim is to show that $Tf=p$.
Since $Tf_k\rightarrow p$ (in $l_1$), we have that
$$\sum_{n=1}^{\infty}|f_k(x_n)-p_n|\rightarrow 0$$
as $k$ goes to $\infty$.
This means that for any $n\in\mathbb{N}$ we have
$$\lim_{k\to\infty}f_k(x_n)=p_n.$$
On the other hand, as $f_k\rightarrow f$, we have that
$$\lim_{k\to\infty}f_k(x_n)=f(x_n).$$
This means that $f(x_n)=p_n$, which implies that $Tf=p$, which is what we needed.
Now, the boundedness of $T$ implies that there is a constant $C$ such that
$$\|Tf\|_{l_1}\le C\|f\|$$
which was to be proved.

$\textbf{Comment}.$ We are tempted to say that the series
$$\sum_{n=1}^{\infty}x_n$$
are absolutely convergent, and afterwards, that there is some $x\in X$ which equals this sum (in which case the problem's constraint will be fulfilled). This however, needs not to be true (and the obstacle is not only that $X$ is not required to be Banach space).
Here is a (counter)example. Let $X=c_0$ and $x_n=e_n$ - the basis unit vectors. Then $(c_0)^*=l_1$ and for any $f\in l_1$ we have
$$\sum_{n=1}^{\infty}|f(e_n)|=\|f\|_{l_1}$$
so that the problem's constraint is satisfied. However, the series
$$\sum_{n=1}^{\infty}e_n$$
are not convergent.
Nevertheless, the problem's conclusion is correct, and the constant is $1$.

Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$ Proof. The function $$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$ is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that $$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$ If we make change of variables in the latter integral $t\to 1-t$ we obtain $$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$ On the other hand, using integration by parts, we obtain $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$ Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$ Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$ On the ot
Read more

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$ $\textbf{
Read more

A problem proposed by prof. Babev

Consider the sequence $$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$ It could be proven that $$\lim_{n\to \infty}n a_n=\frac{1}{2}$$  Now we are going to prove that $$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$ and show how to derive all such limits. $\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as $$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$ $$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$ Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to $$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$ Since for $s\in[0,1]$, $\displaystyle \frac
Read more