$M=\{ (x_1-x_2, x_2-x_3,....)| (x_n)_{n \in \mathbb{N}} \in \ell^{\infty} \}$ is closed.

Question

Consider $M=\{ (x_1-x_2, x_2-x_3,....)| (x_n)_{n \in \mathbb{N}} \in \ell^{\infty} \}$. Show that $M$ is closed in $\ell^{\infty}$. I'm trying to show that any convergent sequence in $M$ converges in $M$.

Take $(x^k)_{k \in \mathbb{N}} $ in $M$ such that $x^k \to x$. $\forall \epsilon > 0 \exists k_0 \in \mathbb{N} $ such that

$$||x^k-x||< \epsilon \text{ whenever } k>k_0$$

I can not conclude that $x \in M$

I'm trying to solve the problem of the image. I do not need M to be closed to apply the version of Hahn Banach that he uses in the tip?

Answer 1

Long comment.

(I). If $A$ is a vector subspace of the normed linear space $X$ and $f$ is a linear functional on $A$ with $\sup \{|f(a):\|a\|=1\}=1$ then $f$ extends uniquely to a functional $\bar f$ on $\bar A$ such that $\sup \{|\bar f(\bar a)|: \|\bar a\|=1\}=1: $ Show that if $(a_n)_n$ and $(a'_n)_n$ are any sequences in $A$ converging to $\bar a\in \bar A,$ then $(f(a_n))_n$ and $(f(a'_n))_n$ converge, and converge to the same value, which we define as $\bar f(\bar a).$ We can then extend $\bar f$ to $f'\in X^*$ with $\|f'\|=1$ by Hahn-Banach applied to $\bar f$.

(II). In the Q, any $y=(y_n)_{n\in \Bbb N}$ is in $M$ iff $y$ has bounded partial sums, i.e. $\infty>\sup \{|\sum_{j=1}^ny_j|: n\in \Bbb N\}.$ As has been shown in the A by Yanko, $M$ is not closed.

(III). A variant of the proof of 2.53 is to take $M'$ to be the vector sub-space of those members of $l_{\infty}$ that have Cesaro Means, and for $x=(x_n)_{n\in \Bbb N}\in M'$ let $L(x)=\lim_{n\to \infty}\frac {1}{n}\sum_{j=1}^n x_j.$ And extend $L$ to a member of $l^*_{\infty}$ with $\|L\|=1.$

(IV). There are uncountably many $L$ satisfying 2.53. They are called generalized limits.

In particular, there exists a binary sequence $y\in l_{\infty}$ \ $\overline {M'}$ with $\|y\|=d(y,\overline {M'})=1,$ and if you look at the details of the proof of the Hahn-Banach theorem, you will see there are uncountably many consistent possible choices for $L(y)$ when extending $L$ from $\overline {M'}$ to the vector space generated by $\{y\}\cup \overline {M'}.$

Answer 2

This is wrong. $M$ is not closed.

First we note that the sequence $y$ with $y_n = \frac{1}{n}$ is not in $M$. Indeed, if $x=(x_n)$ is a sequence such that such that $y_n = x_n - x_{n+1}$ then $x_{n+1} = x_n -\frac{1}{n}$. Inductively this means that $x_{n+1}=-\sum_{k=1}^n \frac{1}{k}$ which diverge, but then $x\not\in l^\infty$.

Now, try to construct for every $k$ a sequence $x^k$ such that the sequence $y^k = (x_1^k-x_2^k,...,x_i^k-x_2^k,...)$ satisfies that $y^k=\frac{1}{n^{1+\frac{1}{k}}}$. This time it is possible because $\sum_{i=1}^n \frac{1}{i^{1+\frac{1}{k}}}$ convergence and so bounded.

I leave the construction of $y^k$ as an exercise. Once you've done, we have that $y^k\rightarrow y$ in $l^\infty$ as $\frac{1}{n^{1+\frac{1}{k}}}\rightarrow\frac{1}{n}$ as $k\rightarrow\infty$ uniformly$^*$.

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$^*$ The point is that $\frac{1}{n^{1+\frac{1}{k}}}-\frac{1}{n}\rightarrow 0$ as $n\rightarrow\infty$ and so point-wise convergence as $k\rightarrow\infty$ implies uniform convergence (as for any $\varepsilon>0$ you only have to deal with finitely many $n$).