# Banach Limit: understanding this step in the proof

In chapter 7 of the book "Functional Analysis, Spectral Theory and Application" by Einsiedler where the discussion concerns Banach Limits. The author proves the statement $$\liminf_{n \rightarrow \infty} \leq \text{LIM}((a_n)) \leq \limsup_{n \rightarrow \infty}$$ along the following lines:

Let $I = \inf_{n \geq 1}a_{n}$ and $S = \sup_{n \geq 1}a_{n}$, so that $|a_n-\frac{I+S}{2}| \leq \frac{S-T}{2}$ for all $n\geq 1$, and hence $$|\text{LIM}((a_n))-\frac{I+S}{2}| \leq \frac{S-I}{2}$$

It is not clear me to how the last statement follows. Here the symbol $\text{LIM}$ denotes the functional $\text{LIM}:\ell^{\infty} \rightarrow \mathbb{R}$ such that $$\text{LIM}((a_n))=L(a_1,\frac{a_1+a_2}{2},...)$$ and $L:\ell^{\infty} \rightarrow \mathbb{R}$ is the functional extended via the Hahn Banach Theorem from $f:c \rightarrow \mathbb{R}:(a_n)\mapsto \lim_{n \rightarrow \infty} a_n$ and so $\text{LIM}$ is the Banach Limit functional. The problem is that since we used Hahn Banach we have no information on the quantity $\text{LIM}((a_n))$ if $(a_n)$ is not convergent in some sense. Indeed we know that $| \text{LIM}((a_n)) | \leq ||(a_n)||_{\infty}$ but how does the above statement follows is still not clear to me.

Could anyone provide some clarification? Thanks!

Banach limits are shift invariant $LIM(\{a_n\})=LIM(\{a_{n+1}\})$, and at the same time bounded linear functional, with respect to $\|\{a_n\}\|=\sup |a_n|$, with norm equal to one.

So $$|LIM (\{a_n\})|\le \sup_{n\in\mathbb N}|a_{n+k}|,$$ for all $k\in\mathbb N$. Thus $$|LIM (\{a_n\})|\le \limsup_{n\in\mathbb N}|a_{n}|,$$ Say that $a_n+c\ge 0$, for all $n$. Then $$LIM(\{a_n\})+c=LIM(\{a_n+c\})\le \limsup |a_n+c|=\limsup (a_n+c)=c+\limsup a_n$$ and hence $$LIM(\{a_n\})\le \limsup a_n.$$ Similarly, if we put $\{-a_n\}$ in the place of $\{a_n\}$, we obtain $$LIM(\{a_n\})\ge \liminf a_n.$$

• Why the proof of LIM$(a_n)\leq \limsup a_n$ by $a_n+c$ but not just $a_n$? I feel like LIM$(a_n)\leq \limsup a_n$ is clear since the Hahn-Banach theorem – Bob Nov 28 '19 at 6:12

The crucial thing you need is $LIM(a_n) \geq 0$ if $a_n \geq 0$ for all $n$. To see this let $0 \leq a_n \leq M$ for all $n$. Then $LIM (M-a_n) \leq M$ because the norm of the sequence $(M-a_n)$ does not exceed $M$. Hence $LIM (M) -LIM (a_n) \leq M$. [ Here $(M)$ is the constant sequence $(M,M,...)$]. But $LIM (M) =M$ so $M-LIM(a_n) \leq M$ which gived $LIM(a_n) \geq 0$. This implies that if $a \leq a_n \leq b$ for all $n$ then $a \leq LIM(a_n) \leq b$. Just take $a=\frac {I+S} 2 -\frac {S-I} 2$ and $b=\frac {I+S} 2 +\frac {S-I} 2$ to complete the argument.

• In that case do you know why the author would not begin with $\inf_{n \geq 1} = I \leq a_{n} \leq S = \sup_{n \geq 1}$? I mean after all he wanted to end up with the statement $I \leq \text{LIM}((a_n)) \leq S$ anyway (which he did). That convoluted argument makes no sense if it follows that way. – Meagain May 19 '18 at 11:49
• @Meagain I proved that $LIM(A_n)$ lies between infimum and supremum of $(a_n)$ . (That is the specific question you asked). However, to get $\liminf$ adn $\limsup$ as the bounds you need the fact that $LIM(a_1,a_2,...)=LIM(a_2,a_3,...)$. Once you know this you can iterate this $k$ times and get $\liminf$ and $\limsup$ by taking limit as $k \to \infty$. – Kavi Rama Murthy May 19 '18 at 12:46
• No I mean your proof was perfect and I think it is the way to go. What I meant was that I could not see the motivation in the author's choice of steps, which I thought to be convoluted. – Meagain May 19 '18 at 12:56
• @Meagain Let me also tell you how to prove translation invariance of LIM. Let $b_n=a_n -a_{n+1}$. Then $\frac {b_1+b_2+...+b_n} n =\frac {a_1-a_{n+1}} n$. Since $a_n$ is bounded we see that $b_n \to 0$. Hence $LIM \{b_n\}=0$ which gives $LIM \{a_1,a_2,...\}=LIM \{a_2,a_3,...\}$. – Kavi Rama Murthy May 19 '18 at 22:43