Continuous operators on $C[0,1] $ are not separable

Question

Taking class in functional analysis, I was told that continuous operators $$A:C[0,1]\rightarrow C[0,1]$$ are not separable. I'm trying to prove it by finding uncountable family of operators $A_\alpha$ such that $$\forall A_1, A_2\in A_\alpha \ \ \ \ \|A_1-A_2\|= \sup\limits_{\|f\|\le 1} \|(A_1-A_2)f\|>\varepsilon$$ where $\varepsilon$ is fixed. Operators of multiplying by constant totally does not make sense, I thought about multiplying by function, but didn't figure it out. So, currently I'm fighting with an example of such family of operators, any hint appreciated!

Answer 1

For $t \in [0,1]$ define $A_t : C([0,1]) \to C([0,1])$ as $$(A_t f)(x) = f(t)x, \qquad f \in C([0,1]).$$ Then $A_t$ is clearly bounded with $\|A_t\| = 1.$

For $s,t \in [0,1], s <t$ consider $f \in C([0,1])$ given by $$f(x) = \begin{cases} 0, &\text{ if } x \in [0,s],\\ \frac{x-s}{t-s}, &\text{ if } x \in [s,t],\\ 1, &\text{ if } x \in [t,1].\\ \end{cases}$$ Then we have $$\|A_t-A_s\| \ge \frac{\|(f(t)-f(s))x\|_\infty}{\|f\|_\infty} = 1.$$

Answer 2

This is an idea.

Take for example $$A_\alpha f(x):=\mathbf 1_{[0,\alpha]}(x)f(x),$$ for $\alpha\in(0,1)$, so it is an uncountable family. This is not technically correct, however, because $A_\alpha f$ is not continuous.

Is there a way to make this example work? I don't know

Answer 3

For $0\le x\le 1$ let $$A_xf=f(x)1$$ where $1$ denotes the constant function. Then $\|A_x\|=1$ and $$\|A_x-A_y\|=2,\quad x\neq y$$ as there is a function (piecewise linear, at most $3$ segments) $f_{x,y}\in C[0,1]$ such that $\|f_{x,y}\|_\infty =1$ and $f_{x,y}(x)=-1,$ $f_{x,y}(y)=1.$

Remark The main reason for the lack of separability is that the dual space of $C[0,1]$ is not separable. Point evaluations are elements of the dual space.