# if $h$ is twice differentiable, what is the largest set on which $|h|$ is twice differentiable?

Let $$h:\mathbb R\to\mathbb R$$ be twice differentiable.

What is the largest set on which $$|h|$$ is twice differentiable?

By the chain rule, $$|h|$$ is differentiable at $$a$$ with $$|h|'(a)=h'(a)\operatorname{sgn}h(a)\tag1$$ for all $$a\in\{h\ne0\}$$. Moreover, if $$a\in\{h'=0\}$$, then $$\left|\frac{|h|(a+t)-|h|(a)}t\right|\le\left|\frac{h(a+t)-h(a)}t\right|\xrightarrow{t\to0}0\tag2$$ by the reverse triangle inequaly and hence $$|h|$$ is differentiable at $$a$$ with $$|h|'(a)=0=h'(a)\operatorname{sgn}h(a)\tag3.$$

So, $$|h|$$ is differentiable at least on $$D_1:=\{h\ne0\}\cup\{h'=0\}$$ with derivative $$h'\operatorname{sgn}h$$. (Can we enlarge $$D_1$$?)

Turning to the second derivative: Using that $$\operatorname{sgn}h$$ is differentiable at $$a$$ with $$(\operatorname{sgn}h)'(a)=0\tag4$$ for all $$a\in\{h\ne0\}\cup\{h=0\}^\circ$$, we obtain (by the chain rule, again) that $$|h|$$ is twice differentiable at $$a$$ with $$|h|''(a)=h''(a)\operatorname{sgn}h(a)\tag5$$ for all $$a\in D_2:=\{h\ne0\}\cup\{h=0\}^\circ$$ (noting that $$\{h=0\}^\circ\subseteq\{h'=0\}$$).

So, $$|h|'$$ is differentiable at least on $$D_2$$. Can we enlarge $$D_2$$?

On the other hand, we can show that $$|h|'$$ is differentiable at $$a$$ with $$|h|''(a)=|h''(a)|$$ for all $$a\in D_3:=\{h=0\}\cap\{h'=0\}\cap(\{h''\ne0\}\cup\{h''=0\}\cap D_1^\circ).$$ However, I'm struggling to see how $$D_2$$ and $$D_3$$ are related and hence whether the latter yields an enlargement of $$D_2$$.

EDIT: And as a third option, it's possible to show that $$|h|$$ is twice differentiable on $$D_4:=\mathbb R\setminus\overline{N'}$$, where $$N':=\left\{a\in\mathbb R:a\text{ is an isolated point of }\left\{h=0\right\}\right\}$$; see revision 3 of this answer (and the comments below): https://math.stackexchange.com/a/3210082/47771.

I'm really struggling to see which result is the strongest.

• Isn't $\{h=0\}^o\subseteq \{h'=0\}$? Commented Jul 1, 2019 at 8:53
• @HagenvonEitzen Sure, you're right. Commented Jul 1, 2019 at 11:36

Can we enlarge $$D_1$$? That is, can $$|h|'(a)$$ exist at points $$a$$ where $$h(a)=0$$ and $$h'(a)\ne 0$$? From $$h(a)=0$$ and $$h'(a)\ne 0$$, it follows that for $$\epsilon>0$$ small enough, we have $$h(x)>0$$ on $$(a,a+\epsilon)$$ and $$h(x)<0$$ on $$(a-\epsilon,a)$$ or vice versa. It follows that the one-sided limit of $$\frac{h(x)-h(a)}{x-a}$$ is $$h'(a)$$ on one side and $$-h'(a)$$ on the other, hence $$|h|'(a)$$ cannot exist.

Can we enlarge $$D_2, D_3, D_4$$? We can show

• $$|h|''(x)=\operatorname{sgn}(h(x))h''(x)$$ on $$\{h\ne0\}$$
• $$|h|''(x)=|h''(x)|$$ on $$\{h=0\}\cap \{h'=0\}\cap \{h''\ne 0\}$$
• $$|h|''(x)=0$$ on $$(\{h=0\}\cap\{h'=0\}\cap\{h''=0\})\setminus \overline{\{h=0\}\cap\{h'\ne0\}}$$

The union of these cases is the maximal set where $$|h|''$$. This follows because on $$\{h=0\}\cap\{h'\ne 0\}$$ not even $$|h|'$$ exists and the definition of derivative of a function at a point requires the function to be defined in an open neighbourhood of that point. If we loosen the usual definition of $$f'$$ to taking only the limit of $$\frac{f(x+h)-f(x)}{h}$$ where $$x+h$$ happens to be in the domain of $$f$$ and that $$x$$ is not an isolated point of the domain, then we can replace the third bullet point above with

• $$|h|''(x)=0$$ on $$\{h=0\}\cap\{h'=0\}\cap\{h''=0\}.$$
• Thank you for your remark, but I'd like to stick to to the ordinary definition of differentiability. Commented Jul 1, 2019 at 11:43
• How is the maximal set (the union of your cases) related to $D_2=\{h\ne0\}\cup\{h=0\}^\circ$? Is the former really larger? Commented Jul 1, 2019 at 11:44