This question is motivated from GRE math subject test.

Let $f$ and $g$ be functions of a real variable such that for all $x$, $$g(x)=\int\limits_{0}^{x}f(y)(y-x)\,\mathrm{d}y\,.$$ If $g$ is three times continuously differentiable what is the greatest integer $n$ for which $f$ must be $n$ times continuously differentiable?

$$ \text{(a) } 1\quad\text{(b) } 2\quad\text{(c) } 3\quad\text{(d) } 4\quad\text{(e) } 5$$

What I thought is to try differentiating the function. By the fundamental theorem of calculus I thought that $\mathrm{d}g/\mathrm{d}x$ should be $f(x)(x-x)=0$ which I guess is wrong so I figured that FTC must not work for multi-variable functions. So I have no idea what to do. Does anyone have any ideas?
Thanks in advance

  • $\begingroup$ Just saying that you're missing an assumption, that $f$ is continuous. $g$ can be smooth without $f$ being continuous. $\endgroup$
    – user223391
    Oct 9, 2016 at 14:32
  • $\begingroup$ Is the answer 2 $\endgroup$ Oct 9, 2016 at 14:42

1 Answer 1


First of all if $f(x)=\chi_{\{0\}}(x)$ then $g\equiv 0$ which is smooth so we must assume $f$ is continuous.

Note: $g(x)=\int_0^x f(y)(y)\ dy-x\int_0^x f(y)\ dy$

Then differentiate by FTC:

$g'(x)=f(x)x-xf(x) -\int_0^x f(y)\ dy=-\int_0^xf(y)\ dy$

You can differentiate again by FTC:


Then to differentiate again we need $f$ to be $C^1$ so the answer is $1$.


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