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Let $t\ge 0$ and suppose some Hilbert space $H$ is isometrically isomorphic to $C_F[0,t]$ (continuous maps $[0,t] \to F$, with the supremum norm) where $F=\mathbb{C}$ or $\mathbb{R}$. I am asked to determine $t$ and $H$ up to isometric isomorphism.

I don't know much about Hilbert spaces, and the only thing I can find in my book is that an infinite-dimensional separable Hilbert space over $F$ is isometrically isomorphic to $\ell^2_F$. I'm not seeing how/if this is useful.

Any help would be appreciated!

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Here's a hint: If $t>0$, it's not too hard to find two functions $f,g\in C_F([0,t])$ such that $\|f\|=\|g\|=\|f+g\|=1$. Show that this is impossible in Hilbert space.

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  • $\begingroup$ Great answer! For OP: the answer to the posed questions are hidden in this answer ;) $\endgroup$ – Cameron Williams Nov 10 '17 at 16:46
  • $\begingroup$ So for example $f(x)=x/t$ and $g(x)=1-x/t$, then $||f||=||g||=||f+g||=1$.. Okay, suppose $T:C_F[0,t]\to H$ is an isometric isomorphism. I'm thinking about using the parallelogram rule $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$. In our case, $||f+g||^2+||f-g||^2=2\ne4=2(||f||^2+||g||^2)$. But I need this inequality in $H$, not in $C_F[0,t]$.. Is $T$ linear? The definition in my book doesn't say anything about that. $\endgroup$ – user316769 Nov 12 '17 at 17:30
  • $\begingroup$ @Insertnamehere $T$ is an isometric isomorphism, so it is linear. $\endgroup$ – Aweygan Nov 12 '17 at 17:42
  • $\begingroup$ My book's definition is: "If $X$ and $Y$ are normed linear spaces and $T$ is an isometry from $X$ onto $Y$ then $T$ is called an isometric isomorphism". But then again the book defines an isometry as a linear map $T$ with $||T(x)||=||x||$ for all $x$. Either way, I get it now. Thanks! $\endgroup$ – user316769 Nov 12 '17 at 17:58
  • $\begingroup$ @Insertnamehere I can see where the confusion comes from. Most texts do not mention linearity in the definition of an isometry. In any case, you're welcome. Glad to help! $\endgroup$ – Aweygan Nov 12 '17 at 18:02

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