# Why is $x^2\sin(1/x)$ not strictly differentiable?

I am on Wikipedia reading on strict differentiability and I don't particularly understand the example proving a function that is differentiable does not have to be strictly differentiable, where $$f(x)=x^2 \sin(1/x)$$, $$f(0)=0$$. How can we show this is differentiable directly using difference quotients. Also then how can we show it's not strictly differentiable using our difference quotients?

The simplest setting in which strict differentiability can be considered, is that of a real-valued function defined on an interval $$I$$ of the real line. The function $$f: I \rightarrow \mathbf{R}$$ is said strictly differentiable in a point $$a \in I$$ if $$\lim _{(x, y) \rightarrow(a, a)} \frac{f(x)-f(y)}{x-y}$$ exists, where $$(x, y) \rightarrow(a, a)$$ is to be considered as limit in $$\mathbf{R}^{2}$$, and of course requiring $$x \neq y$$. A strictly differentiable function is obviously differentiable, but the converse is wrong, as can be seen from the counter-example $$f(x)=x^{2}$$ sin $$\frac{1}{x}, f(0)=0, x_{n}=\frac{1}{\left(n+\frac{1}{2}\right) \pi}, y_{n}=x_{n+1}$$. One has however the equivalence of strict differentiability on an interval $$I$$, and being of differentiability class $$C^{1}(I)$$. (Transcribed from Wikipedia screenshot)

so far I have;

To show $$f(x)$$ is differentiable, I would assume I'd use the statement that if $$x^2$$ is differentiable and $$\sin(1/x)$$ is differentiable (which can be shown by the difference quotients, then $$x^2 \sin(1/x)$$ is also differentiable?

• $f'$ is not continuous at $0$, let $y \to x$ before $x\to 0$ you'll have $\lim_{x\to 0}f'(x)$ Commented Dec 28, 2019 at 22:18
• What don't you understand exactly? Have you tried finding the limit along the sequence mentioned? Commented Dec 28, 2019 at 22:21
• I don't know how to find the limit along the sequence, do I take f(x) as $x^2 sin(1/x))$, but then what is f(y)? and is our x and y in this case, $x_n$ and $y_n$, if so what is our a? Commented Dec 28, 2019 at 23:04
– zhw.
Commented Dec 29, 2019 at 3:34
• @zhw.: the given image contains a definition of strictly differentiable, but yes this should have been explicitly given in the post. A function $f:I\to\mathbb{R}$ defined on an interval $I$ is said to be strictly differentiable at $a\in I$ if the two variable limit $$\lim_{(x, y) \to (a, a)} \frac{f(x) - f(y)} {x-y}$$ exists. Commented Dec 29, 2019 at 3:58

Recall negation of function limit definition.

$$\text{if } \exists p_k, q_k\in\mathbb{R^n}: \lim_{k\rightarrow \infty} p_k, q_k=x_0: \lim_{{k\rightarrow \infty}} g(p_k) \neq \lim_{{k\rightarrow \infty}} g(q_k) \Rightarrow \text{no limit}$$

Let us use this fact to prove that the desired limit doesn't exist.

$$\begin{multline} \exists p_k=\left(\begin{smallmatrix}x_{2k}\\y_{2k}\end{smallmatrix}\right), q_k=\left(\begin{smallmatrix}x_{2k+1}\\y_{2k+1}\end{smallmatrix}\right) \in \mathbb{R}^2: \lim_{k\rightarrow \infty} p_k, q_k= \left(\begin{smallmatrix}0\\0\end{smallmatrix}\right):\\ \frac{2}{\pi}=\lim_{{k\rightarrow \infty}} \frac{f(x_{2k})-f(y_{2k})}{x_{2k}-y_{2k}} \neq \lim_{{k\rightarrow \infty}} \frac{f(x_{2k+1})-f(y_{2k+1})}{x_{2k+1}-y_{2k+1}}=-\frac{2}{\pi} \end{multline}$$

where $$f(z) = z^2\sin\frac{1}{z}$$, $$x_n=\frac{1}{(n+1/2)pi}$$ and $$y_n=x_{n+1}$$.

\begin{aligned} \lim_{x \rightarrow 0\\y \rightarrow 0}\cfrac{x^2\sin\frac{1}{x} - y^2\sin\frac{1}{y}}{x-y} &=\lim_{n \rightarrow \infty}\cfrac{x_n^2\sin\frac{1}{x_n} - y_n^2\sin\frac{1}{y_n}}{x_n-y_n}\\ &= \lim_{n \rightarrow \infty}\cfrac{\left(\frac{1}{(n+1/2)\pi}\right)^2\sin(n+1/2)\pi - \left(\frac{1}{(n+3/2)\pi}\right)^2\sin(n+3/2)\pi}{\frac{1}{(n+1/2)\pi}- \frac{1}{(n+3/2)\pi}}\\ &= \lim_{n \rightarrow \infty}\cfrac{\left(\frac{1}{(n+1/2)\pi}\right)^2(-1)^n - \left(\frac{1}{(n+3/2)\pi}\right)^2(-1)^{n+1}}{\frac{1}{(n+1/2)(n+3/2)\pi}}\\ &= \lim_{n \rightarrow \infty}\left[\left(\frac{1}{(n+1/2)\pi}\right)^2 + \left(\frac{1}{(n+3/2)\pi}\right)^2\right](-1)^{n}(n+1/2)(n+3/2)\pi\\ &= \lim_{n \rightarrow \infty}\frac{2n^2+4n+5/2}{(n+1/2)^2(n+3/2)^2\pi^2}(-1)^{n}(n+1/2)(n+3/2)\pi\\ &= \lim_{n \rightarrow \infty}\frac{n^2+2n+5/4}{(n+1/2)(n+3/2)}\frac{(-2)^{n}}{\pi}\\ \end{aligned}

It is clear to see, that

\begin{aligned} n = 2k &\rightarrow p_k \rightarrow\lim_{n \rightarrow \infty} \cfrac{\sim 2n^2}{\sim \pi n^2} = \frac{2}{\pi}\\ n = 2k+1 &\rightarrow q_k \rightarrow\lim_{n \rightarrow \infty} -\cfrac{\sim 2n^2}{\sim \pi n^2} = -\frac{2}{\pi} \end{aligned}

To sum up, $$f$$ is not strictly differentiable.

More general result: Suppose $$f$$ is differentiable on an open interval $$I,$$ and $$f'$$ is discontinuous at some $$x_0\in I.$$ Then $$f$$ is not strictly differentiable on $$I.$$

Proof: Let $$a_n$$ be a sequence of distinct points converging to $$x_0.$$ By the MVT there exist $$b_n$$ between $$x_0$$ and $$a_n$$ such that

$$\frac{f(a_n)-f(x_0)}{a_n-x_0}=f'(b_n).$$

As $$n\to \infty,$$ we see $$f'(b_n) \to f'(x_0).$$ On the other hand, because $$f'$$ is discontinuous at $$x_0,$$ there exists a sequence $$c_n\to x_0$$ such that $$f'(c_n)$$ does not converge t0 $$f'(x_0).$$

Now find sequences $$s_n,t_n\to x_0$$ such that

$$\left | \frac{f(s_n)-f(t_n)}{s_n-t_n}-f'(b_n) \right |<1/n$$

and sequences $$u_n,v_n\to x_0$$ such that

$$\left | \frac{f(u_n)-f(v_n)}{u_n-v_n}-f'(c_n) \right |<1/n.$$

Both the sequences $$(s_n,t_n),(u_n,v_n)$$ converge to $$(x_0,x_0).$$ Consider the sequence $$(x_n,y_n)$$ obtained by putting these two sequences together. I.e., $$(x_n,y_n)$$ is the sequence

$$(s_1,t_1),(u_1,v_1),(s_2,t_2),(u_2,v_2), \dots$$

Then $$(x_n,y_n)\to (x_0,x_0)$$ and the corresponding difference quotients converge to $$f'(x_0)$$ through odd values of $$n,$$ but fail to converge to $$f'(x_0)$$ through even values of $$n.$$

Therefore $$f$$ is not strictly differentiable on $$I.$$

In our specific problem $$f'$$ is not continuous at $$0,$$ so this $$f$$ is not stricly differentiable on $$\mathbb R$$ by the general result.

• "Because $f'$ is discontinuous at $x_0,$ there exist $L\ne M$ and sequences $a_n,b_n\to x_0$ such that $f'(a_n) \to L$ and $f'(b_n) \to M$": why is this? It's not true for discontinuous functions in general. So if it's true for derivatives (and I don't know that it is), it is a special result that can't just be assumed to be true. Commented Dec 29, 2019 at 20:42
• @TonyK That's a good point. I edited by answer.
– zhw.
Commented Dec 30, 2019 at 1:07
• You haven't addressed my point at all. Commented Dec 30, 2019 at 1:11
• @TonyK It's a shame you only get one downvote. I think I addressed your "true for derivatives" bit by referring to the well known Darboux property. What else do I need to address?
– zhw.
Commented Dec 30, 2019 at 1:20
• Why does the Darboux property imply what you claim that it implies? You say "From this it follows...", and I don't see that at all. Can you try to explain it please? Commented Dec 30, 2019 at 1:42