5
$\begingroup$

Let $l_{\mathbb{R}}^{\infty}$ be the linear space of all real-valued bounded sequences over $\mathbb{R}$ for $x=(x_n)_n$ define $$ p(x) = \lim_{n \to \infty} \sup \frac{x_1 + \dots x_n}{n} \\ W = \{x \in l_{\mathbb{R}}^{\infty} : \lim_{n\to \infty} \frac{x_1 + \dots + x_n}{n} \text{ exists } \} $$

Prove that there is a linear functional LIM on $l_{\mathbb{R}}^{\infty}$ with the following properties

a) $LIM(x_1,x_2,\dots) = LIM(x_2,x_3,\dots)$

b) $\lim \inf_{n \to \infty} x_n \leq LIM(x_1,x_2, \dots) \leq \lim \sup_{n \to \infty} x_n$

c) LIM is continuous with $\|LIM\|=1$.

Since $W$ is a linear subspace I defined a functional $\psi_0(x) = \lim_{n \to \infty} \frac{x_1 + \dots + x_n}{n}$ where $\psi_0(x) = p(x)$ on $W$.

Then by Hahn-Banach there exits $\psi$ s.t $\psi(x) = \psi_0 (x)$ on $W$ and $\psi(x) \leq p(x)$ on $V$. But how can I continue?

$\endgroup$

1 Answer 1

6
$\begingroup$

First, the following easy corollary of Hahn-Banach for normed spaces might be useful:

Proposition: Let $X$ be a normed space, and let $Y$ be a subspace with $\phi\in Y^*$. Then $\phi$ has an extension to $\tilde{\phi}\in X^*$ such that $\|\phi\|=\|\tilde{\phi}\|$.

To prove this, apply the usual Hahn Banach theorem with $p(x)=\|\phi\|\|x\|$

Now, as you did before, define

$\psi_0:W\to\mathbb{R}$ by

$\psi_0((x_n)_n)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}x_n$

By Hahn-Banach, we can extend this to a function $Lim:l^{\infty}\to\mathbb{R}$ such that

$\|Lim\|=\|\psi_0\|$

Let's prove part c) first now. In light of the above, we need only show that

$\|\psi_0\|=1$. Indeed, we have $\frac{1}{n}\sum_{k=1}^{n}x_n\leq \frac{n\|(x_n)_n\|_{\infty}}{n}=\|(x_n)_n\|_{\infty}$

Thus $\|\psi_0\|\leq 1$. To see that $\|\psi_o\|=1$, consider the sequence $x\in W$ with only ones as its entries.

This proves c).

For part a)

Note that by linearity:

$Lim(x_1,x_2,...)=Lim(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)$

$=\psi_0(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)=Lim(x_2,x_3,x_4,...)$

This proves a)

Finally, for part b), let $x=(x_n)\in l^{\infty}$ Choose $a,b$ so that

$a<\liminf_{n\to\infty}(x_n)$ and

$b>\limsup_{n\to\infty}(x_n)$

Then there is $N$ such that for $n>N$, we have $a<x_n<b$.

Denote now 1 by the sequence with only ones as its entries and denote $y=(y_n)_n$ as the sequence such that $y_n=x_{n+N}$

Then $w:=y-a\cdot$1 and $z:=b\cdot$1$-y$ are positive and bounded sequences.

Thus,

$0\leq Lim(w)=Lim(y)-a=Lim(x)-a$

So that, $a\leq Lim(x)$

where the last equality was obtained using the shift invariance proven in a)

Similarly, we obtain $Lim(x)\leq b$.

Since, $a$ and $b$ were arbitrary, b) follows.

The one step that needs justification is that if $x=(x_n)_n$ is positive and bounded, then $Lim(x)\geq 0$

Indeed, without loss of generality, $\|x\|\leq1$(otherwise, rescale x)

Now, $1-Lim(x)=Lim($1$-x)\leq\|$1$-x\|\leq 1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .