My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)

Let $f$ and $g$ be continuous functions over the reals s.t.

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

and $g$ is three times continuously differentiable,

what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?

my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?

(feel free to edit the title to best fit my question)


We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound: $$g(x)=\int_0^xf(y)ydy-x\int_0^xf(y)dy$$ Taking the derivative with respect to $x$, and using $$\frac {d}{dx}\int_0^x w(t)dt=w(x)$$ we have $$\frac{dg(x)}{dx}=f(x)x-xf(x)-\int_0^xf(y)dy=-\int_0^xf(y)dy$$ Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.

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  • $\begingroup$ got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$? $\endgroup$ – Hossien Sahebjame Jan 8 '19 at 21:45
  • 1
    $\begingroup$ You are correct. $\endgroup$ – Andrei Jan 8 '19 at 21:47

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