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Are $(\ell^\infty(\mathbb{Z}))^*$ and $ (\ell^\infty(\mathbb{N}))^*$ isomorphs?

I think that I could establish the next function $\Phi:(\ell^\infty(\mathbb{N}))^*\to(\ell^\infty(\mathbb{Z}))^*$ such that for all $g\in (\ell^\infty(\mathbb{N}))^*$, $\Phi_g(x)=\Phi_g((x_n)_{n\in\mathbb{Z}})=g((x_n)_{n\in\mathbb{N}})$ and $0$ if $n\in\{...,-1,0\}$. Is correct the function?

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    $\begingroup$ I believe there is a natural isomorphic isomorphism between $\ell^{\infty}(\mathbb{Z})$ and $\ell^{\infty}(\mathbb{N})$ using a bijection between $\mathbb{Z}$ and $\mathbb{N}$. $\endgroup$ – Floris Claassens Mar 19 at 12:20
  • $\begingroup$ Dear Jose, welcome to Mathematics Stackexchange. We value questions that provide some context and show some effort from the OP. If you could say why you think that these spaces are isomorphic (or not) and what you tried to prove (or disprove) this, that could considerably increase the quality of your question as perceived by the community. $\endgroup$ – MaoWao Mar 19 at 13:49
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Let $a:\mathbb{N}\rightarrow\mathbb{Z}$ be a bijection, we claim that $$\hat{a}:\ell^{\infty}(\mathbb{Z})\rightarrow \ell^{\infty}(\mathbb{N}),f\mapsto f\circ a$$ is an isomorphic isomorphism. First of all note that $$\hat{a}^{-1}:\ell^{\infty}(\mathbb{N})\rightarrow \ell^{\infty}(\mathbb{Z}),f\mapsto f\circ a^{-1}$$ is the inverse of $\hat{a}$, so $\hat{a}$ is bijective. Clearly it is also linear, so $\hat{a}$ is an isomorphism.

Finally let $f\in \ell^{\infty}(\mathbb{Z})$ and note that $$\|f\|_{\infty}=\sup_{z\in\mathbb{Z}}|f(z)|=\sup_{n\in\mathbb{N}}|f(a(n))|=\|\hat{a}(f)\|_{\infty}$$ so $\hat{a}$ is isomorphic.

As $\ell^{\infty}(\mathbb{Z})$ and $\ell^{\infty}(\mathbb{N})$ are isomorphic isomorph their dual spaces coincide.

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  • $\begingroup$ Thank you so much! $\endgroup$ – Jose Esparrago Mar 20 at 10:51

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