# Pseudoconverse of “Differentiability of Product” in Analysis

It is well known that if $f$,$g$ are both differentiable functions, then $fg$ is also a differentiable function. However, we know that the converse is not true. Take for example $f = |x|$ and $g = |x|$. Both functions are nondifferentiable, while their product $fg = x^2$ is clearly differentiable. Take another example $f = \theta(x)$ and $g = 1- \theta(x)$. Then $fg = 0$ is differentiable, but $f,g$ are not.

What I find interesting in this case and many other examples is that the converse seems almost true, in the sense that when $fg$ is differentiable, $f,g$ seems to behave quite nicely. In fact, they seem to be differentiable in the sense of distributions. In the case of $f = |x|$, we can schematically write $f'(x)= -1 + 2 \theta(x)$ with the understanding that this expression only makes sense when $f'$ is integrated against some other continuous function in $x$. In the case of $f = \theta(x)$. $f'(x) = \delta(x)$ which is a distribution that also makes sense only under integration.

So my question is: can we formulate a pseudoconverse to the "differentiability of product" in analysis? In particular, what, if anything, can we say about $f,g$ individually if we know that $fg$ is a standard differentiable function (and assuming that neither of $f,g$ is identically zero)? Can we say more if we demand $f,g$ to be continuous? Are the nice examples I provided just coincidences? If so, could someone provide some pathological examples to counter my expectations?

• what function is $\theta$ ? oh wait it is the Heaviside function? – mathreadler Jul 29 '17 at 21:34
• Indeed your examples are only coincidences... Also, you really should think of distributions as something bigger than $L^1_{\rm loc}$. – John B Jul 29 '17 at 21:48
• @JohnB Thanks! Could you please write a full answer? – Zhengyan Shi Jul 29 '17 at 21:50
• @mathreadler Yes indeed. – Zhengyan Shi Jul 29 '17 at 21:50
• @TsemoAristide. Good point. See edits. – Zhengyan Shi Jul 29 '17 at 21:51

Well, let $$f(x)=\left\{\begin{array}{ll} 1 & x\in\mathbb{Q}\\ -1 & \mbox{else} \end{array}\right.$$

And $$g(x)=-f(x)$$ Then, clearly $f,g$ are not differentiable - actually, not even continuous - but $$(fg)(x)=f(x)g(x)=-f^2(x)=-1$$ which is obviously a differentiable function. So, the differentiability of $fg$ does not give many valuable information about $f,g$.

Moreover, let $h:I\to\mathbb{R}$ be a positive differentiable function. We will write $h$ in the form of $fg$ where none of $f,g$ is differentiable. For this puprpose, let us consider: $$f(x)=\left\{\begin{array}{ll} \sqrt{h(x)} & x\in\mathbb{Q}\\ -\sqrt{h(x)} & \mbox{else} \end{array}\right.$$ and let $$g(x)=f(x)$$ Then, it is again clear that $$(fg)(x)=f(x)g(x)=f^2(x)=h(x)$$ Also, since $h$ is s positive, it is clear that $f,g$ are discontinuous, so, non-differentiable.

• Although they are pointwise very bad functions, their almost everywhere behaviour is nice and that is what is seen by the distributional definition of the derivative. – Calvin Khor Jul 30 '17 at 11:29
• @CalvinKhor, using a nonmeasurable set you have very bad functions even in the distributional sense. – Martín-Blas Pérez Pinilla Jul 30 '17 at 16:09