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$X=C[0,1]$ define $T:X\to X$ by $T(f(x))=\int_{0}^{x} f(t) dt$

Then I need to find whether it is one and onto.

If $T(f(x))=0$ then $\int_{0}^{x} f(t) dt=0$ taking derivative we get $f(x)=0$ so injective .

suppose for $g(x)\in X$ we have $f(x)\in X$ such that $T(f)=g$ so $\int_{0}^{x} f(t) dt=g\Rightarrow f=g'$ but I dont know whether $g'$ exist so $T$ is not onto, am I right?

Myshkin
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    What hits constants? – Alex Youcis Sep 12 '13 at 10:23
  • excuse me for my previous comment : what i mean is that $g$ is continuous does not imply $g$ is differentiable :P –  Sep 12 '13 at 10:35
  • It seems the following. You are right and each function from $TX$ is differentiable, while there exist continuous and non-differentiable functions on the segment $[0;1]$ (for instance, put $f(x)=x$ for all $0\le x\le 1/2$ and $f(x)=1-x$ for all $1/2\le x\le 1$). Moreover, $g(0)=0$ for each function $g\in TX$. So, the map $T$ is not onto. – Alex Ravsky Sep 17 '13 at 03:05

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You correctly demonstrated that $T$ is injective.

The easiest way to see that $T$ is not surjective is to observe that $Tf(0)=0$ for every $f$. Hence, the constant function $1$ is not in the range of $T$. (As Alex Ravsky said)

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