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Let

$$ \ f(x) = \begin{cases} x^{2}\sin(\frac{1}{x}), & \text{if $x \not = 0$} \\ 0, & \text{if $x = 0$} \end{cases} $$

and $$g(x) = \begin{cases} 0, &\text{if } x\not \in \mathbb Q \\ \frac{1}{b}, &\text{if } x = \frac{a}{b} \ \text{ with } a \in \mathbb Z, \ b \in \mathbb Z^{+} \text{ and } \gcd(a,b) = 1 \end{cases}$$ Let $h(x) = f(g(x))$

Determine where $h(x)$ is continuous and differentiable.

Here is my trial.. firstly derive $$ h(x) = \begin{cases} \frac{1}{b^{2}} \sin(b) & \text{where } x = \frac{a}{b}, \\ 0 & \text{where } x \not \in \mathbb Q \end{cases}$$

Then we can show $h(x)$ is not continuous at every $x \in \mathbb Q $ as following: Assume $p = \frac{a}{b}$ then $\ f(p) = f(\frac{a}{b}) = \frac{1}{b^{2}}\sin(b)$, but given any irrational sequence $x_{n}$ converging to p} \$ we have $\lim_{n \to \infty}f(x_{n}) = 0$ since $x_{n}$ is irrational, then we proved the discontinuity in $\mathbb Q$.

But how can I prove that $h$ is continuos for $x \not \in \mathbb Q$ ? Hope someone can help...

1 Answers 1

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To prove $h$ is continuous for irrational numbers we have to prove that $$\forall x \notin \Bbb Q\ \forall \epsilon > 0\ \exists \delta > 0 : |x-y| < \delta \Rightarrow f(y) < \epsilon$$ If we take an interval of length $1$ around $x$, we see there are at most $3$ fractions that can be written $\frac{a}{2}$ in this interval (two such fractions have a distance of $\frac{1}{2}$ between them). Similarly, $\forall n \in \Bbb N $, there is a finite number of fractions in this interval that can be written $\frac{a}{n}$. But we observe $h(\frac{a}{n}) \leqslant \frac{1}{n^2}$, which means if we take $n$ large enough, we can assure $h(\frac{a}{b}) < \epsilon \ \forall b > n$. As there are only a finite amount of fractions where the denominator is smaller than n ( in the chosen interval), there is a minimum distance between $x$ and such a fraction. If we take $\delta$ smaller than this distance, the condition to be continuous is satisfied.

I cannot completely prove there exists an irrtional number where $h$ is not differentiable it exists, here is why:

Take any rational point $a_1$. $h(a_1) \neq 0$ (unless $a_1 = 0$). We now define the interval $I_1 = [a_1-h(a_1),a_1+h(a_1)]$. As $\Bbb Q$ is dense in $\Bbb R$, we know there is another rational point in $I_1$, call it $a_2$.Now define the interval $I_2 = [a_2-h(a_2),a_2+h(a_2)]$.The intersection of $I_1$ and $I_2$ is non-empty, and we find another rational point, $a_3$, in this intersection.We continue, and as the length of these intervals tend to $0$, we know $\lim_{n\to \infty}a_n$ exists. Call it $x$. Now $h$ is clearly not differentiable in $x$, as the derivate would be $1$ for any $a_1$ and $0$ for any irrational number. There is no reason $x$ should be rational, that is why I think there exist irrational numbers where $h$ is not differentiable.

This does not exclude that in some places the function is differentiable (with derivative $0$).

I hope this helps.