Can we always multiply some function that is not differentiable everywhere with function that is to obtain differentiable product?

Question

First of all, I think that before stating the general question it would be okay to make some concrete example of what do I have in mind.

Let us take the function $f(x)=|x|$.

We could write this function as $f(x)= \begin{cases}x&{x>0}\\-x&x<0 \\ 0& {x=0} \end{cases}$.

This function is differentiable everywhere except at the point $x=0$.

Now let us multiply this function with, say, function $g(x)=x$.

Now we have $f(x)g(x)=\begin{cases}{x^2}&{x>0}\\{-x^2}&{x<0}\\ 0&{x=0} \end{cases}$.

Clearly $f(x)g(x)$ is differentiable everywhere.

What did we do here?

We multiplied function which was not differentiable at one point with some other function which is of class $C^{\infty}$ and we obtained function which is everywhere differentiable.

Now I would like to ask the question which deals with the general case:

Suppose that we have some real function of a real variable $f$ that is continuous on some set $(a,b)$ and that is not differentiable on some subset of $(a,b)$ (the subset could be only one point as in the above described example or it could be the whole set $(a,b)$ so that we have everywhere continuous but nowhere differentiable function). Could it be that there always exists some function $g$ (which could depend on $f$) of class $C^{\infty}$ (function that is infinitely times differentiable) which is not the zero function (so $g$ is not the function $g(x)=0$) and which is such that we have that the function $fg$ (the product of the functions $f$ and $g$) is differentiable on the set $(a,b)$?

Answer 1

No, we cannot. Let $f \colon (a,b) \to \mathbf R$ a continous, nowhere differentiable function. Suppose, $g \in C^\infty(a,b)$ is given with $fg$ being differentiable everywhere. We will show $g = 0$. Let $x \in (a,b)$, then, as $fg$ and $g$ are differentiable at $x$, if we had $g(x) \ne 0$, the quotient $f = \frac{fg}g$, defined on some neighbourhood of $x$, would also be differentiable at $x$ by the quotient rule. Hence, $g(x) = 0$. As $x$ was arbitrary, $g = 0$.

Answer 2

This is cheating a little, but if the points of nondifferentiability are isolated, you could form a function which vanishes in a neighborhood of each of those points, and is not identically zero, and multiply by that.