Does there exist a linear functional $L$ on $l^{\infty}$ such that $L(x)=\lim_{n\to \infty} x_n$?

Question

Does exist a linear functional $L$ on $l^{\infty}$ such that $L(x)=\lim_{n\to \infty} x_n$ if it exists?

I did following: I chose $x=(1,1,1,\dots) \in l^{\infty}$ and $y=(-1,-1,-1,\dots) \in l^{\infty}$

Obviously, $x+y=(0,0,\dots)$ and $L(0)=L(x+y)=L(x)+L(y)=\lim_{n\to \infty} x_n+y_n$ but right hand side doesn't have limit and LHS has limit which is $0$. So, I a little bit confused

Start with $L(x)=\lim_{n\to \infty} x_n$ for convergent sequences and then apply Hahn-Banach Theorem. — geetha290krm, May 04 '23 at 07:34
If you do not care about continuity of $L$ then HB theorem is not needed. — Ryszard Szwarc, May 04 '23 at 07:56
There is no contradiction: $L(0)=0$, $L(x)=1$, $L(y)=-1$, $L(x+y)=L(0)=0$. Maybe you wanted to take $x$ alternating $x=(+1,-1,+1,-1,...)$ and $y=-x$. — daw, May 04 '23 at 11:47
this might answer your questoion: https://math.stackexchange.com/questions/996263/extension-of-the-limit-operator-on-l-infty — daw, May 04 '23 at 11:49
Thank you for suggestion, I defined bounded linear functional from $c \to \mathbb R$ then extend it to $l^{\infty}$ — Elise9, May 04 '23 at 14:00

score 1 · Accepted Answer · answered May 05 '23 at 16:57

1

Limit operator is bounded linear functional on $c$. And rest comes from Hahn-Banach Theorem.

answered May 05 '23 at 16:57

nozalp10

108

Does there exist a linear functional $L$ on $l^{\infty}$ such that $L(x)=\lim_{n\to \infty} x_n$?

1 Answers1