f and g are function of a real variable such that $g(x) = \int_0^x f(y)(y-x)dy$ for all $x$. If g is three times continuously differentiable, what is the greatest integer n for which f must be $n$ times continuously differentiable. Apparently the answer is $1$. Can anyone explain how to do this problem?

Since the derivative of $g$ is $0$, the integral must be constant, but I don't see how to quickly use this to say anything about $f$.

  • $\begingroup$ Surely they assumed $f$ was continuous? $\endgroup$
    – zhw.
    Aug 2, 2015 at 22:25
  • $\begingroup$ @zhw. They didn't assume that, they just said that they were functions and that g was 3 times continuously differentiable. $\endgroup$ Aug 3, 2015 at 11:40
  • $\begingroup$ what if you consider as f the dirichlet function? then g is identically zero but f is not even continuous $\endgroup$
    – user264117
    Aug 24, 2015 at 16:49

2 Answers 2


By Fundamental Theorem of Calculus we can compute $g'(x),g''(x),g'''(x)$ directly:




  • $\begingroup$ Thanks, I see what I did wrong. $\endgroup$ Aug 2, 2015 at 20:57
  • $\begingroup$ @agha If you allow me to suggest a correction, I would remark that the derivative (w.r.t $x$) of $\int_{0}^{x}f(y)\;dy$ is $f(x)$, NOT $(f(x)-f(0))$. So your computation should be $g'(x)=f(x)x-\int_{0}^{x}f(y)\;dy-x(f(x))=-\int_{0}^{x}f(y)\;dy$ and $g''(x)=-f(x)$. $\endgroup$
    – Ramiro
    Aug 2, 2015 at 22:09
  • $\begingroup$ @Ramiro Guerreiro, thank you. I've corrected the answer. $\endgroup$
    – agha
    Aug 2, 2015 at 22:17
  • $\begingroup$ Sorry but how did you get the $f(x)x$ and $-xf(x)$ in the first derivative? $\endgroup$
    – lsy
    Jun 23, 2016 at 3:33
  • $\begingroup$ Can someone explain the first derivative? Exactly which formula is used? $\endgroup$
    – lsy
    Jun 23, 2016 at 4:50

Write $g(x) = -x\displaystyle \int_{0}^x f(y)dy+\displaystyle \int_{0}^x yf(y)dy\to g'(x) =-\displaystyle \int_{0}^x f(y)dy-xf(x)+xf(x)=-\displaystyle \int_{0}^x f(y)dy\to g''(x) = - f(x)\to g'''(x) = -f'(x)$ From this we see that $n = 1$


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