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I would like to know how I prove that the map $$T:C([0,1])\rightarrow C^1([0,1])$$ $$(Tf)(s):=\int_0^sf(t)dt$$ is well-defined.

I suspect that proving $Ran(T)\subset C^1([0,1])$ would be enough. Is this true?

For this I would start with an $f\in Ran(T)$, i.e. of the form $f(s)=(Tg)(s)=\int_0^sg(t)dt$ for some $g\in C([0,1])$. How do I prove this is continuously differentiable though?

Help would be very welcome.

azureai
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  • For $T$ to be well defined you need to show that $Tf$ is a continuously differentiable function for any continuous $f$. I.e. $Tf\in C^1[0,1]$ for any $f\in C[0,1]$. But this is implied by the fundamental theorem of calculus as the answers below indicate. – T. Eskin Apr 13 '16 at 16:42

2 Answers2

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The fundamental theorem of calculus tells you that

$$ f'(s) = (Tg)' (s) = g(s). $$

John Hughes
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  • Does this make the function well-defined, too? – azureai Apr 13 '16 at 16:31
  • Yes, it does. You might want to look at this q/a for a better understanding of what "well-defined" actually means: http://math.stackexchange.com/questions/606917/well-defined-function-what-does-it-mean – John Hughes Apr 13 '16 at 18:08
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Well, there are theorems which tell you when $s\mapsto \int_0^s f(t) dt $ is differentiable. For continuous $f$ this happens to be true, and the derivative is known to be $$ \frac{d}{ds}\int_0^s f(t) dt = f(s)$$ which is continuous by assumption.

Thomas
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