Given the Banach space $C[0,1]$={$f$:$[0,1]$$\longrightarrow$$\mathbb{C}$, continuous} equipped with sup norm,
I would like to find a bounded linear operator $T$ : $C[0,1]$$\longrightarrow$$C[0,1]$ that is surjective but not injective.
And all I can think of is a differential operator, but not all continuous functions are differentiable.
So a differential operator wouldn't work unless the domain is restricted.
I have come up with $T(f(x))=f(\frac{x}{2})-f(1-\frac{x}{2})$
T is not injective, but I am not convinced if T is even surjective.
So any help would be appreciated.
Thank you.
$T: C[0, 1] \to C[0, 1]$ defined by $Tf(x) = f(x/2)$ is an example.
For any $g \in C[0, 1]$, $T^{-1}(g)$ is the set of all $f \in C[0, 1]$ satisfying $$ f(x) = g(2x) \text{ for } 0 \le x \le \frac 12 $$ and that are infinitely many.
More generally, as demonstrated in functions between function spaces, for any continuous $\varphi: [0, 1] \to [0, 1]$ the mapping $T$ defined by $Tf(x) = f(\varphi(x))$ is surjective but not injective exactly if $\varphi$ is injective but not surjective.