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By the Banach-Mazur theorem, every separable Banach space $X$ embeds into $C([0,1])$. When $X$ is reflexive, it is not possible to find a sequence of disjointly supported, non-negative functions in any isometric image of $X$ in because this would generate a copy of $c_0$. I am intersted in some special families of functions and (im)possibility of their containment in reflexive subspaces of $X$.

Let $X$ be a separable Hilbert space. Is it possible to embed $X$ isometrically into $C([0,1])$ in such a way that the image of $X$ contains a sequence of distinct strictly increasing functions that map 0 to 0 and 1 to 1?

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  • $\begingroup$ Isomorphic or isometric embedding? Isomorphic is possible. $\endgroup$
    – Bill Johnson
    Commented Nov 19, 2019 at 21:38
  • $\begingroup$ @BillJohnson, I am mostly interested in the isometric case, but I am curious about the isomorphic too. $\endgroup$
    – A. U.
    Commented Nov 20, 2019 at 7:30
  • $\begingroup$ If $f_n$ are $C^1$ functions that vanish at $0$ and at $1$ s.t. the sup norm of their derivatives is less than $1/2$, then $t+f_n(t)$ are increasing functions. Every infinite dimensional Banach space can be isomorphically embedded into $C[0,1]$ so as to contain such a sequence that is linearly independent. $\endgroup$
    – Bill Johnson
    Commented Nov 21, 2019 at 0:15

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This is impossible. Each such function has norm 1 and only one supporting functional (point value at 1) in $C[0,1]$ so a'fortiori in this Hilbert space. But in Hilbert space the supporting functional uniquely defines the element.

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