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Is it possible to define an isometry from l∞ to C[0,1] using disjoint functions?

I can define an isometry from l-p space to L-p space using disjoint functions but I am having difficulties with l∞ to C[0,1].

taupi
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1 Answers1

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Such an isometry does not exist, because $C[0,1]$ is separable and $l^\infty$ is not (however, an isometry would preserve this property).

Severin Schraven
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  • Let $\alpha_n:=1-\frac{1}{n}$ for $n\in \mathbb{N}$. What about the operator $T$ mapping $(a_n){n\in \mathbb{N}}\in \ell^{\infty}$ to the function $x\mapsto \sum{n=1}^{\infty}L(n)\chi_{[\alpha_n,\alpha_{n+1})}$, where $L(n):\mathbb{R}\to \mathbb{R}$ is the linear function interpolating the points $(\alpha_n,a_{n})$ and $(\alpha_{n+1},a_{n+1})$? Isn't this an isometry? EDIT: Nevermind, I got it, if $\lim_{n\to \infty}a_n$ does not exist then the image function is cannot be defined in the endpoint $x=1$. Interesting. – Lorenzo Q May 05 '18 at 11:05
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    @LorenzoQ. Indeed, your operator gives you an isometry from $l^\infty$ to $C_b[0,1)$. This makes sense, as we have no separability in this case. – Severin Schraven May 05 '18 at 12:57