# Two non-differentiable functions whose product is differentiable.

So I was wondering while studying analysis if there is any case where two functions aren't differential at $0$ (kind of like $1/x$) but is differentiable at 0 when combined (i.e. $fg$).

I mean this for functions that are defined on $\mathbb{R}$.

• Don't worry about the "wondering." Even if your question can be shot down rather quickly this is a natural mathematical activity. You can spot a mathematician by the fact that his/her wastebasket is filled with wild conjectures and fruitless attempts, but there on the desk, is a solitary short paper with a correct conjecture. Commented Nov 15, 2015 at 4:41

## 3 Answers

Take $f(x)=|x|$ and $g(x)=|x|$.

• oh of course, what about the case where only one function is differentiable at 0 though? Commented Nov 14, 2015 at 19:11
• $f$ and $g$ are both not differentiable at $0$. If you want an example where only one is not differentiable, take $f$ as above, and $g(x)=0$. Commented Nov 14, 2015 at 19:13
• Ok...what about a function that is differentiable at zero but not at any point? Is this possible? Commented Nov 15, 2015 at 22:02
• @SelfStudy that's a complete different question, maybe you should ask a new one. Commented Nov 15, 2015 at 22:37
• Wow, this line of questioning looks suspiciously like Exercise 5.2.2 on p.152 of Abbott, Understanding Analysis, 2nd Edition. Commented Oct 24, 2016 at 12:30

Let $h :R\to R$ be positive and differentiable at every point. Example 1.Let $f:R\to R$ be positive and nowhere continuous. For example let $f(x)=1$ when $x$ is rational and $f(x)=2$ when $x$ is rational. Example 2. Let $f:R\to R$ be continuous ,positive, and nowhere differentiable.(Such functions do exist.)...In both examples let $g(x)=h(x)/f(x)$. In both examples , neither $f$ nor $g$ are differentiable anywhere but their product is differentiable everywhere. In example 2., both $f$ and $g$ are continuous. For a third example suppose $f(x)=0$ for $x\leq 0$ and that $f$ is discontinuous on $[0,\infty)$. Let $g(x)=f(-x)$.

On $\mathbb R$ let $f$ be the indicator of $[0,\infty)$ and $g$ the indicator of $(-\infty,0)$. Their product is $0$, whereas each of the functions is not differentiable at $0$.

Furthermore, you dont need your functions to be defined on $\mathbb R$ to say something about differentiability in $0$, since differentiability is a local property.