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If $f'$ is piecewise smooth on $[a,b]$, then $f$ is piecewise smooth on $[a,b]$.

Is this statement true?

I know that if we let $f'$ be piecewise smooth on $[a,b]$. Then, by definition, both $f'$ and $f''$ are piecewise continuous on $[a,b]$. Now, what I am left to show is that $\int f'$ is piecewise continuous on $[a,b]$.

In other words, is the anti-derivative of a piecewise continuous function also piecewise continuous?

Thanks in advance!

  • 0
    Is "smooth" defined to mean infinitely differentiable? No matter how it's defined, your problem is not equivalent to showing that anti-derivatives of piecewise continuous functions are piecewise continuous-the latter property is much weaker than smoothness.2012-10-27
  • 0
    The book I'm using defines piecewise smoothness of a function $f$ on $[a,b]$ as $f$ and $f'$ being piecewise continuous on $[a,b]$.2012-10-27
  • 0
    In this problem, if I can show $\int f'=f$ is piecewise continuous, then, because $f'$ is piecewise continuous, $f$ is piecewise smooth.2012-10-27
  • 1
    Assume $f$ is not continuous at some point $x$, what can you say about $f'$?2012-10-27

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