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

Solution 1:

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.

Solution 2:

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.