Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 11E / Solution

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.