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 $\mathbb{R}$.
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$\begingroup$ 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. $\endgroup$– B. S. ThomsonCommented Nov 15, 2015 at 4:41
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3 Answers
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Take $f(x)=|x|$ and $g(x)=|x|$.
answered Nov 14, 2015 at 18:13
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$\begingroup$ oh of course, what about the case where only one function is differentiable at 0 though? $\endgroup$ Commented Nov 14, 2015 at 19:11
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2$\begingroup$ $f$ and $g$ are both not differentiable at $0$. If you want an example where only one is not differentiable, take $f$ as above, and $g(x)=0$. $\endgroup$ Commented Nov 14, 2015 at 19:13
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$\begingroup$ Ok...what about a function that is differentiable at zero but not at any point? Is this possible? $\endgroup$ Commented Nov 15, 2015 at 22:02
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$\begingroup$ @SelfStudy that's a complete different question, maybe you should ask a new one. $\endgroup$ Commented Nov 15, 2015 at 22:37
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1$\begingroup$ Wow, this line of questioning looks suspiciously like Exercise 5.2.2 on p.152 of Abbott, Understanding Analysis, 2nd Edition. $\endgroup$ Commented Oct 24, 2016 at 12:30
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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 $g(x)=f(-x)$.
answered Nov 14, 2015 at 23:39
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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.