On $p(x) = \limsup_{n\to\infty} \frac{x_1+x_2+\dots+x_n}{n}$

Question

The following is an Exercise from Bruckner's Real Analysis:

Define a function p on the sequence space l^∞ by $$p(x) = > \limsup_{n\to\infty} \frac{x_1+x_2+\dots+x_n}{n},$$ and define a linear functional l on the subspace c by $$l(x) = \lim_{n→∞} x_n$$. (a) Show that p is subadditive and positively homogeneous on $l^∞$.
(b) Apply the Hahn–Banach theorem to obtain a linear functional L on $l^∞$ such that, for $x={\{x_n}\}$,

(i) $L(x)≥0$ if $x_n≥0$ for all n∈N.

(ii) $L({\{x_1 ,x_2 ,x_3,...}\})=L({\{x_2,x_3 ,x_4 ,...}\})$ for all $x ∈ l^∞$.

(iii) $\liminf x_n ≤ L(x) ≤ \limsup x_n$ for all $x ∈ l^∞$.

(iv)$L(x)=\lim_{n→∞} x_n$ for all x∈c. Thus L provides a notion of limit applied to all bounded sequences. The four properties (i) through (iv) are ones that we would expect of a generalized limit. One calls L a Banach limit.

(c) Calculate $L({\{0, 1, 0, 1,...}\})$.

item (a) : $p(ax)=ap(x)$ is obvious, but how $p(x+y) \le p(x)+p(y)$?

item (b) : I have no idea even for start.

item (c) : $L$ is an extension of $l$ so how the limit of ${\{0, 1, 0, 1,...}\}$ can exit at all to calculate?

The answer here is not helpful at all!

Answer 1

(a). Let $x, y \in \ell_\infty$. For each $n$, define $n$, $X_n = \frac{x_1 + \cdots + x_n} {n} $ and $Y_n = \frac{y_1 + \cdots + y_n} {n} $. So, we have

$$p(x+y) = \limsup_n (X_n+Y_n) \le \limsup_n X_n + \limsup_n Y_n =p(x)+p(y)$$

(b) $c$ is a linear subspace of $\ell_\infty$. Define $l$ in $c$ by, for all $x \in c$, $ l(x) = \lim_n x_n$. Then, $$ l(x) = \lim_n x_n = \limsup_n \frac{x_1 + \cdots + x_n} {n} = p(x) $$ (Note that $\lim_n x_n = \limsup_n \frac{x_1 + \cdots + x_n} {n}$ because $x$ is a convergente sequence)

So we can apply Theorem 12.28 (Hahn-Banach), and get a functional $L$ defined on $\ell_\infty$ that extends $l$ and satisfies, for all $x \in \ell_\infty$, $ L(x) \leq p(x)$.

Item $(iv)$: Since $L$ extends $l$, item $(iv)$ is trivial.

Item $(iii)$: Now, we have, for all $x \in \ell_\infty$, $$ L(x) \leq p(x) = \limsup_n \frac{x_1 + \cdots + x_n} {n} \leq \limsup_n x_n$$ Since $L$ is linear, we also have $$ -L(x) =L(-x) \leq \limsup_n -x_n = - \liminf_n x_n$$ that means $$ L(x) \geq \liminf_n x_n$$ So we have proved item $(iii)$: $ \liminf_n x_n \leq L(X) \leq \limsup_n x_n$, for all $x \in \ell_\infty$.

Item $(i)$: it is an immediate consequence of item $(iii)$. In fact, for any $x \in \ell_\infty$, if $x =\{x_1,x_2, \cdots \}$ and $x_n \geq 0$, for all $n$, then we have that $0 \leq \liminf_n x_n \leq L(x)$.

Item $(ii)$ First, note that given any $x, y \in \ell_\infty$, $L(x) - L(y) = L(x-y) \leq p(x-y)$ and $L(y) - L(x) = L(y-x) \leq p(y-x)$. So $|L(x) - L(y)| \leq \max \{p(x-y), p(y-x)\}$.

Now, for any $x \in \ell_\infty$, if $x =\{x_1,x_2, \cdots \}$, let $y$ be the sequence $y =\{x_2, \cdots \}$. It is easy to see that $$p(x-y) = \limsup_n \frac{x_1 - x_2 + \cdots + x_n -x_{n+1}} {n} = \limsup_n \frac{x_1 -x_{n+1}} {n} = 0$$ and $$p(y-x) = \limsup_n \frac{x_2 - x_1 + \cdots + x_{n+1}- x_n} {n} = \limsup_n \frac{x_{n+1} - x_1} {n} = 0$$ So, we have that $|L(x) - L(y)| =0$. It means $L(x) = L(y)$.

(c) Let $x = \{ 1, 0, 1, 0, \cdots \}$ and $y = \{ 0, 1, 0, 1, \cdots \}$ . By (b) item $(ii)$ , $L(x)= L(y)$. We know that $$ L(x) +L(y) = L(x+y) = L({1,1,1 \cdots}) = 1$$ ($ L({1,1,1 \cdots}) = 1$ from (b) item $(iv)$)

So $L(y) = \frac{1}{2}$.