Let $\Lambda$ be any Banach limit on $\ell^\infty$, where $\ell^\infty$ denotes the space of bounded real sequences. A Banach limit is defined as a linear functional $\Lambda$ such that $$ \Lambda(\tau x)=\Lambda(x), \forall x\in\ell^\infty$$ $$ \liminf_{n\rightarrow\infty}x_n\leq\Lambda(x)\leq\limsup_{n\rightarrow\infty}x_n$$ where we write $x=(x_n)_{n\in\mathbb{N}}$ for a sequence $x\in\ell^\infty$ and we define left translation on $\ell^\infty$ by $(\tau x)_n=x_{n+1},n=1,2,\dots$.

I would like to show that $\Lambda\in(\ell^\infty)^*$, which means that $\Lambda$ is a continuous, linear functional on $\ell^\infty$. Thus I need to show that $\Lambda$ is continuous. How do I do this?

Furthermore I wish to show that there exists a continuous, linear functional $\Lambda_0\in(\ell^\infty/c_0)^*$ such that $\Lambda=\Lambda_0\circ q_0$, where $$q_0:\ell^\infty\rightarrow\ell^\infty/c_0$$ is the quotient map and $$c_0=\{(x_n)\in\Lambda^\infty\mid \lim_{n\rightarrow\infty}x_n=0\}$$

I can't seem to get anywhere with these questions. Any help is greatly appreciated.

  • $\begingroup$ Oops, I forgot that, it has been added. Thank you. $\endgroup$ – Kevin Apr 20 '16 at 12:17

That $\Lambda$ is continuous follows directly from the second estimate: $$ -\|x\|_\infty\leq \liminf_{n\to\infty} \Lambda(x_n)\leq \Lambda(x)\leq \limsup_{n\to\infty}\Lambda(x_n)\leq\|x\|_\infty $$ Thus, $|\Lambda(x)|\leq \|x\|_\infty$ for all $x\in \ell^\infty$.

For your second claim define $\Lambda_0(x+c_0):=\Lambda(x)$. This is well-defined since $x-y\in c_0$ implies $\Lambda(x-y)=0$, i.e. $\Lambda(x)=\Lambda(y)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.