# Show that $T$ is isometric and not surjective. Proof validation.

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?

### Problem

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 $$T f = y \ \ \square.$$

• Looks good to me. Dec 19, 2021 at 15:12
• Very nice solution! Dec 19, 2021 at 16:27
• 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 $f(0) = 1$ Dec 19, 2021 at 22:29