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I have to show that there exists a linear functional $T:\ell ^{\infty}\rightarrow \mathbb{R}$ satisfying the following conditions:

  • $\liminf x\le Tx\le \limsup x\ \forall x\in \ell^{\infty}$
  • $T(\tau x)=Tx\ \forall x\in \ell ^{\infty}$

where $(\tau (x))_n=x_{n+1} \forall x\in \ell^{\infty}\ \forall n\ge 1$
An indication was to define the sequence of functionals $T_n(x)=\frac{1}{n}\sum_{k=1}^nx_k\ \forall x\in \ell^{\infty}\ \forall n\ge 1$ and to consider the subspace $F=\{ x\in \ell^{\infty},\ (T_nx)\ converges\}$.
So I defined $T:F\rightarrow \mathbb{R}$ by $Tx=\lim T_nx\ \forall x\in \ell^{\infty}$, I proved that $Tx\le \limsup x\ \forall x\in F$ and I applied the Hahn-Banach theorem to extend $T$ to a functional $T:\ell ^{\infty}\rightarrow \mathbb{R}$ satisfying $Tx\le \limsup x\ \forall x\in \ell^{\infty}$ and I proved that $\liminf x\le Tx\ \forall x\in \ell^{\infty}$.
I couldn't however prove that $T(\tau x)=Tx\ \forall x\in l^{\infty}$ (I proved it $\forall x\in F$ though).
I would appreciate any help you could provide me with. Thanks!

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    https://math.stackexchange.com/questions/2225002/banach-limit-hahn-banach-theorem – daw Nov 09 '19 at 18:13
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    The idea is to use that for $x\in l^\infty$, $\tau x-x \in F$ and $T(\tau x-x )=0$, so $T(x) = T(x-\tau x) + T(\tau x) = T(\tau x) $. – daw Nov 09 '19 at 18:15

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As Daw pointed out in the comments above, it is sufficient to prove that $\forall x\in \ell ^{\infty}\ x-\tau x\in F$ and $T(x-\tau x)=0$ which is quite straighforward:
Let $x\in \ell^{\infty}$ and $n\in \mathbb{N}$. Then $T_n(x-\tau x)=\frac{1}{n}\sum_{k=1}^{n}(x_k-x_{k+1})=\frac{x_1-x_{n+1}}{n}\rightarrow 0$ because $x$ is bounded.
So $x-\tau x\in F$ and $T(x-\tau x)=0$. It follows that $Tx=T \tau x\ \forall x\in \ell^{\infty}$.

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