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

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? 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 $$\{y\}\cup \overline {M'}.$$

• Only time that what I intended to be a comment turned into an Accepted answer. May 14, 2019 at 11:05

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$$^*$$.

$$^*$$ 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 $$n$$).

• I'm trying to solve the problem of the above image. I do not need $M$ to be closed to apply the version of Hahn Banach that he uses in the tip? May 10, 2019 at 21:58
• $x_{n+1}=x_{n} - \frac{1}{n}$, Right? May 10, 2019 at 22:15
• @Lucas It may be that he defines $M$ a little different than you understood. It says all the elements of the form $x-x'$ where $x'$ as above, but you didn't post what's above so I don't know what $x'$ is. May 11, 2019 at 12:19
• Look at item (4) of the statement of the question, is defined $x '$. May 11, 2019 at 17:10
• @Lucas Right you don't need that $M$ is closed. You can consider $\overline{M}$ instead. Since $dist(1,M)=1$ we have that $1\not\in\overline{M}$ and so you can use Hahn-Banach May 11, 2019 at 19:25