Define $Tf(x) = \int_{0}^{x} f(t)dt$ Then, choose the correct option

Question

$(C[0, 1], ||.||_{\infty})$ denotes the set of all real-valued continuous functions on $[0, 1]$ with $||f||_{\infty}$:= $\sup{|f(t)|, t ∈ [0, 1]}$. For each $x \in [0, 1],$ define $Tf(x) = \int_{0}^{x} f(t)dt$

Then, choose the correct option

$1.$ $T$ is injective but not surjective.

$2.$ $T$ is surjective but not injective.

$3.$ $T$ is bijective.

$4.$ $T$ is neither injective nor surjective.

My attempt : i thinks T is bijective take $f(x)= x$ option $3)$ will true

Is its True ?

Any hints/solution will be appreciated

Answer 1

If $g$ is a continuous function then it follows by the fundamental theorem of calculus that $\int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.

A linear map, $T$ is injective if and only if $\ker T = \{0\}$. What can you conclude about $f$ if for every $x$

$$\int_{0}^{x}f(t)dt = 0$$

Hint apply fundamental theorem of calculus

Answer 2

If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $g\in C[0,1]$ is of the form $g=Tf$ for some $f\in C[0,1]$.