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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.

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

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