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Let $f$ be a continuous real valued function defined on a closed interval $[a,b]$. Let $F$ be the function defined, for all $x$ in $[a.b]$ by $$F(x) = \int_{a}^{x} f(t) dt$$.

Does it necessary follow that $F(x)$ is of class $C^\infty$ on the open set $(a,b)$? Or is it simply $C^1$?

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If $f$ isn't differentiable then can $F$ be $C^\infty$? –  anon Mar 8 '13 at 4:35
    
@anon, no (10 char) –  nam Mar 8 '13 at 4:38
    
Actually, $f$ is $C^k$ if and only if $F$ is $C^{k+1}$. –  1015 Mar 8 '13 at 5:03

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up vote 1 down vote accepted

Expanding on @anon's comment, if you suppose $f$ is a continuous non-differentiable function, say $f(x)=|x|$ on the interval $[-1,1]$, then your $F(x)$ is continuously differentiable, that is, $C^1[-1,1]$, but it is not $C^2[-1,1]$ since $f$ is not differentiable.

You can in fact, take a continuous, nowhere differentiable function $f$ and get that $F$ is $C^1$ but not $C^2$ at any point.

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that's what I was thinking, the absolute value. You are speaking about a nowhere differentiable function for $f(t)$ right? –  nam Mar 8 '13 at 4:39
    
Right (for the second part). –  Clayton Mar 8 '13 at 4:40
    
If $f$ is $C^{n-1}$, does it follow that $F$ is $C^{n}$? –  nam Mar 8 '13 at 4:41
    
If $F'=f$ and $f\in C^{n-1}$, what does that say about $F^{(n)}=(F')^{(n-1)}$? –  Clayton Mar 8 '13 at 4:43
    
I am tempted to multiply $(n - 1)^2$ –  nam Mar 8 '13 at 4:52

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