Is $T:X-X$ as $T(f(x))=\int_0^{x}f(t)dt,\; \forall f\in X$.one-one onto?

Question

Let $X=C(0,1)$. Define $T:X-X$ as $T(f(x))=\int_0^{x}f(t)dt,\; \forall f\in X$. Then I have to determine if the mapping is one-one onto or not.

For one-one

$T(f(x))=T(g(x))$

$\Rightarrow \int_0^{x}f(t)dt=\int_0^{x}g(t)dt$

Differentiating with respect to $x$

$\Rightarrow f(x)=g(x)$. Hence one-one.

For onto I am not sure,

$T(f(x))=\int_0^{x}f(t)dt$ according to me means that whatever function we input, we would get something that is dependent on $x$. Hence we won't be able to get any constant functions which do belong in $X$

Is this correct?

Answer 1

No, it is not onto, since, for each $f\in X$, $T(f)(0)=0$. So, for instance, the constant function $1$ does not belong to the range of $T$.

Answer 2

If $g \in T(X)$, then $g(x)=\int_0^{x}f(t)dt$ for some $f \in X$. Since $f$ is continuous, g is differentiable.

Conclusion: if $g \in X$ is not differentiable, then $g \notin T(X).$

We have $T(X)=C^1(0,1)$.