GRE Math Subject Practice Test Question $53$

Question

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 $f$.

Answer 1

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

$$g'(x)=f(x)x-\int_{0}^{x}f(y)\;dy-xf(x)=-\int_{0}^{x}f(y)\;dy$$

$$g''(x)=-f(x)$$

$$g'''(x)=-f'(x)$$

Answer 2

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 $n = 1$