I have a solution to the following problem and I want to ensure that my reasoning is correct. Can you find any error? Would you solve it in a different manner?


Let $S = \{s_1, s_2, ... \}$ be a countable dense set in $[0,1]$. Define the linear map $T: C([0,1]) \to l^{\infty} $ by $T(f) = (f(s_1), f(s_2), ...)$.

  1. Show that $T$ is isometric.
  2. Show that $T$ is not surjective.

Note: $C([0,1])$ is equipped with $|| \cdot ||_{\infty}$

Solution 1

I need to show that $\forall f: ||T(f)|| = ||f||$. Since $S\subset [0,1]$, it is clear that $||T(f)|| \le ||f||$. We need to show that $||T(f)|| \ge ||f||. $Since $f$ is continuous and $[0,1]$ compact, $\exists x\in [0,1]$ such that $|f(x)| = ||f||$. Since $S$ is dense in $[0,1], \exists (x_1, x_2, ....), x_i \in S$ such that $x_i \to x$. Since $f$ is continuous, $f(x_i) \to f(x) \implies |f(x_i)| \to ||f||$, which means that $||T f|| = \sup\limits_{s\in S}\{|f(s)|\}\ge \sup\limits_{i\ge 1}\{|f(x_i)|\} \ge ||f|| \ \ \square$.

Solution 2

We need to find $y \in l^{\infty}$ such that $ y \notin T(C[0,1])$. Let $y = (1,0,0,0,...)$ and assume $\exists f$ such that $T f= y$. That means that $f(s_1) = 1$ and $\forall i \ge 2: f(s_i) = 0$. Since $f$ is continuous, $\exists x \in [0,1]$ such that $f(x)=1/2$. Again, since $S$ is dense, there is a sequence $(x_1, x_2,...), x_i \in S$ such that $x_i \to x \implies f(x_i) \to f(x) = 1/2$. But this is a contradiction, since $f(x_i) \in \{ 0,1 \}$. Hence $\nexists f\in C([0,1]) $ such that $T f = y \ \ \square.$

  • $\begingroup$ Looks good to me. $\endgroup$ Dec 19, 2021 at 15:12
  • $\begingroup$ Very nice solution! $\endgroup$
    – Benjamin
    Dec 19, 2021 at 16:27
  • $\begingroup$ For the second, you took a correct $y$, alternatively you could proceed as follows, take a sequence of $S$, non of its element is 0 but it converges to 0, but then $f(x_i) = 0$ for all $i$, and so should the limit, which doesn't happen, since $f(0) = 1$ $\endgroup$
    – Physor
    Dec 19, 2021 at 22:29


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