Is $T$ surjective,injective or both.

Question

Let $(C[0,1],\|\cdot\|_\infty)$ be the set of continuous real valued functions on $[0,1]$ with $\|f\|_\infty=\sup_{t\in [0,1]} |f(t)|$.

For each $x\in [0,1]$ define $Tf(x)=\int _0^x f(t) $

Then is $T$ surjective,injective or both?

$T$ is a linear functional such that $T:C[0,1]\to C[0,1]$

Now $T$ is injective $\iff \ker T=\{0\}$

$$f\in \ker T\implies Tf=0\implies \int _0^xf(t)=0 \forall x\in [0,1]$$

Take $$F=\int _0^x f(t) \implies F^{'}(x)=f(x)=0\implies f(x)=0$$

Hence $T$ is injective.

For surjective:

I am unable to show this . Please help.

Answer 1

No, $T$ is not surjective, since each element of the range of $T$ is a differentiable functions. So, for instance, the function $t\mapsto\left|x-\frac12\right|$ doesn't belong to the range.

Answer 2

$T$ is not surjective.

Assume there exists $f \in C[0,1]$ such that $Tf \equiv 1$.

Thus $$\int_0^x f(t)\,dt = 1, \forall x \in [0,1]$$

Taking the derivative, we get $f(x) = 0, \forall x \in [0,1]$. But then $Tf \equiv 0$.