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Consider the function $$f(x)=\begin{cases}0&,x\in\mathbb Q\\x^2&, x\in\mathbb R\setminus\mathbb Q\end{cases}$$ Prove that $f$ is differentiable only at $x=0$.

My approach:

Let us take any point $a$ and assume that $f$ is continuous at $a$.

We know that there exists a sequence $\{q_n\}_{n\ge 1}$ of rational numbers such that $$\lim_{n\to\infty}q_n=a.$$

Also we know that there exists a sequence $\{r_n\}_{n\ge 1}$ of irrational numbers such that $$\lim_{n\to\infty}r_n=a.$$

Now since we have assumed that $f$ is continuous at $a$, this implies that $$\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}f(r_n)=f(a)...(1)$$

Now $$\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}0=0$$ and $$\lim_{n\to\infty}f(r_n)=\lim_{n\to\infty}r_n^2=a^2.$$

Therefore by $(1)$ we have $0=a^2\implies a=0.$ Now $f(0)=0$.

This implies that $f$ is continuous only at $x=0$.

Now let us take any sequence $\{x_n\}_{n\ge 1}$ such that $$\lim_{n\to\infty}x_n=0.$$

Now if the limit $$\lim_{n\to\infty}\frac{f(x_n)-f(0)}{x_n-0}=\lim_{n\to\infty}\frac{f(x_n)}{x_n}$$ exists, then we can conclude that $f$ is differentiable at $x=0$.

But, how to systematically show the same?

Sanket Biswas
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  • In this question, if it was asked to find the points at which $f$ is continuous, then the technique that I have used in the first part of the question, would it suffice? To be specific, I have shown that if $f$ is continuous at some point $a$, then $a=0$ and not if $a=0$, then $f$ is continuous at $a$. So, is it correct? – Sanket Biswas Feb 23 '20 at 09:54

3 Answers3

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you have $\dfrac{f(x)}{x}=0$ or $x$, depending on the rationality of $x$. Hence for all $x\neq 0$, we have $\vert \dfrac{f(x)}{x}\vert\leq \vert x\vert$. So the letf hand side goes to zero as $x$ goes to zero and you may conclude that $f$ is differentiable at $0$ and $f'(0)=0$.

GreginGre
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3

There is no need to first study continuity.

You form two sequences converging to some $x_\infty$ by the irrationals and by the rationals.

We first assume $x_\infty$ irrational, and

$$\frac{f(x_n)-f(x_\infty)}{x_n-x_\infty}=\begin{cases}\dfrac{x_n^2-x_\infty^2}{x_n-x_\infty}\to2\,x_\infty&\text{ for }x_n\notin\mathbb Q\\\dfrac{-x_\infty^2}{x_n-x_\infty}\text{ d.n.e.}&\text{ for }x_n\in\mathbb Q.\end{cases}$$

Next, with $x_\infty$ nonzero rational,

$$\frac{f(x_n)-f(x_\infty)}{x_n-x_\infty}=\begin{cases}\dfrac{x_n^2}{x_n-x_\infty}\text{ d.n.e.}&\text{ for }x_n\notin\mathbb Q\\\dfrac{0}{x_n-x_\infty}\to0&\text{ for }x_n\in\mathbb Q.\end{cases}$$

Finally, with $x_\infty=0$,

$$\frac{f(x_n)-f(x_\infty)}{x_n-x_\infty}=\begin{cases}\dfrac{x_n^2}{x_n}\to0&\text{ for }x_n\notin\mathbb Q\\\dfrac{0}{x_n}\to0&\text{ for }x_n\in\mathbb Q.\end{cases}$$ and the derivative is $0$.

2

We have $$ \frac{f(x_n)}{x_n} = \frac{1}{x_n} \begin{cases} 0, & x_n \in \mathbb{Q} \\ x_n^2, & x_n \notin \mathbb{Q} \end{cases} = \begin{cases} 0, & x_n \in \mathbb{Q} \\ x_n, & x_n \notin \mathbb{Q} \end{cases} \overset{n \to \infty}{\longrightarrow} 0$$ as both terms converge to $0$.

blat
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