# Continuously differentiable composition/convolution.

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.
• 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 $n = 1$? – Hossien Sahebjame Jan 8 '19 at 21:45