Let $V=C[0,1]$, continuous real valued function on $[0,1]$
Consider the linear transformation $$T(f)=\int_{0}^{x}f(t)dt$$
Is $T$ onto?
Attempts
What does onto mean? So for each $g\in V$ we have some $f\in V$ such that $T(f)=g$. In other words,
$$\int_{0}^{x}f(t)dt=g(x)$$ Taking derivative on both sides(Using Leibniz Rule here) we get $f(x)=g'(x)$.
Take any function $g$ whose derivative does not exist in $[0,1]$ and we are done!.
Am I right??
Another approach would be to observe that constant function is not in the image of $T$.
If I am missing something please point out.
Yes, you are right.
Another way of proving that $T$ is not onto comes from noticing that you always have $T(f)(0)=0$. So, take a function $g\in C[0,1]$ such that $g(0)\neq0$ and you're done (again).