Relating to the problem here: Show that the derivative of a function is not continuous, i.e. we have a function: $$g(x)=\begin{cases} x+2x^2\sin\left(\frac{1}{x}\right)&\text{ if }x\neq0\\\ 0&\text{ if }x=0 \end{cases}$$ Check that the function g is not of Class $C^1$ in any open interval around $x=0$, the Jacobian matrix of g at $x=0$ is nonsingular/invertible, but g is not injective in any open interval around $x=0$.

My understanding is that if g is of Class $C^1$, then partial derivatives exists and are continuous, and thus g is differentiable. But I'm not sure how to check the above properties of g, any help is appreciated.

  • $\begingroup$ What's $g'(x)$ away from 0? And what's $g'(x)$ as $x \to 0$? $\endgroup$ – Jon Warneke Dec 2 '15 at 23:58
  • $\begingroup$ $$g'(x)=4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)+1 ,\text{ if }x\neq0$$ $\endgroup$ – EmmaJ Dec 3 '15 at 1:14

You are correct in your definition of $C^1$. Just realize that since $g$ is a function of one variable, in this particular case $C'$ simply means $g'$ exists and is continuous, and the Jacobian matrix will be the $1 \times 1$ matrix with entry $g'$. So the Jacobian matrix being nonsingular at $x = 0$ just means that $g'(0) \neq 0$.

Here's a sketch of what you need to do:

  1. Compute $g'(x) = 1 - 2\cos(1/x) + 4x\sin(1/x)$ at $x \neq 0$ using elementary differentiation rules. Show that $g'(0) = 1$ using the difference quotient definition. This shows that the Jacobian matrix of $g$ at $x = 0$ is nonsingular. To show that $g'$ is not continuous, show that $g'$ does not tend to $1$ as $x \to 0$. You could use an $\epsilon, \delta$ proof, but I think a more efficient method is to recall that if $g'$ is continuous at $x = 0$, then for every sequence $\{x_n\} \to 0$, the sequence $\{g'(x_n)\} \to g'(0) = 1$. Hint: when is $\cos(1/x) = 1?$.

  2. For noninjectivity note that $g$ is continuous. So to show that $g$ is not injective on any interval $(-c,c)$, you need only show that $g'$ takes both negative and positive values on any such interval. Hint: any such interval contains zeroes of both $\cos(1/x)$ and $\sin(1/x)$.

  • $\begingroup$ Thanks! I got the first part but having trouble finding the interval where g' takes both negative and positive values, could you explain further why such interval contains zeroes of both cos(1/x) and sin(1/x)? $\endgroup$ – EmmaJ Dec 3 '15 at 4:07
  • $\begingroup$ The zeroes of $\sin(1/x)$ are $\{1/\pi n : n \in \mathbb{Z}\}$ and the zeroes of $\cos(1/x)$ are $\{2/\pi(2n + 1) : n \in \mathbb{Z}\}$. Both of these sets contain elements that are arbitrarily small in absolute value (just pick $n$ with $\vert n \vert$ large enough), so any interval $(-c,c)$ will contain some of these zeroes (infinitely many, in fact). $\endgroup$ – Ethan Alwaise Dec 3 '15 at 5:06
  • $\begingroup$ Also, if we've shown that there is a sequence $\{x_n\}$ with $\{x_n\} \to 0$ as $n$ approaches infinity, such that $g'(x_n)=0\ \forall n$, can we conclude that g' is not continuous and g is not of Class $C^1$ in any open interval around x=0? $\endgroup$ – EmmaJ Dec 3 '15 at 5:18
  • $\begingroup$ Yes. All you need is a single sequence $\{x_n\}$ with $\{x_n\} \to 0$ such that $\{g'(x_n)\}$ does not tend to $1$. $\endgroup$ – Ethan Alwaise Dec 3 '15 at 5:25

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