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Given $T \colon C[0,1] \to C[0,1]$ defined by $$Tx(t):= \int_0^t x(r) dr$$ for each $t\in [0,1]$, where $C[0,1]$ is the normed space of continuous real-valued (or complex-valued) functions defined on the closed interval $[0,1]$ with the norm given by $||x|| \colon= \max_{t \in [0,1]} |x(t)|$.

How to find the range of this operator?

This operator is one-to-one; so an inverse exists. How to find this inverse?

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  • $\begingroup$ The range seems to be the set of all differentiable functions which are zero at zero. if I'm right. It is one to one, inverse is obtained by differentiating. $\endgroup$ – user59671 Feb 16 '13 at 1:16
  • $\begingroup$ Note that $Tx (0) = 0$ for all $x$, hence it cannot have an inverse. If $Tx=y$, then $\dot{y} = x$, $y(0) =0$. $\endgroup$ – copper.hat Feb 16 '13 at 1:17
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Suppose $Tx=y$, we have: $$y(t)=(Tx)(t)= \int_0^t x(r) dr$$ Since $x$ is continuous, $y$ is differentiable and we have: $$x=y'$$ So $$T^{-1}(y)=y'$$

About range of $T$, it's clear that each $Tx$ is differentiable and $$(Tx)(0)=0$$ Let $y\in C[0,1]$ be a differentiable function such that $$y(0)=0$$ we have: $$(Ty')(t)=\int_0^t y'(r) dr=y(t)$$ provided this integral exists. So which integral do you use? e.g if this definition is used then the integral always exists and the range of $T$ is all differentiable functions that are 0 at 0. About Riemann integration I think the range is more limited.

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  • $\begingroup$ Why should this Riemann integral should not exist? I reckon it is equal to $y(t)-y(0)=y(t)-0=y(t)$. What can possibly go wrong, I wonder? $\endgroup$ – Saaqib Mahmood Feb 16 '13 at 12:12
  • $\begingroup$ @SaaqibMahmuud: It doesn't exist always for derivatives. $\endgroup$ – user59671 Feb 17 '13 at 4:02

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