Is there a function $f(x)$ and a point $x=a$ such that $|f(x)|$ differentiable at $x=a$ but $f(x)$ is not differentiable at $x=a$?
Isn’t it the absolute value that is creating sharp points which makes the function not differentiable? (e.g. $|x|$ at $x=0$).
Thanks!
If you allow $f$ to be discontinuous, then you can potentially use $\vert\cdot\vert$ to remove discontinuities. Consider for example the function $$f:\Bbb R \rightarrow \Bbb R, x \mapsto \left \{ \begin{array}{ll} 1 & x\geq 0\\ -1 & x<0\end{array}\right.$$ It clearly is not continuous in 0 but $\vert f \vert$ is a constant function, hence differentiable.
$\vert \cdot \vert$ creating sharp points is only possible, if it is applied to a function, whose output switches signs. Even then it doesn’t have to kill differentiability. For example I think the function $$g:\Bbb R \rightarrow \Bbb R, x \mapsto \vert x^3\vert$$ is differentiable, with $$g‘:\Bbb R \rightarrow \Bbb R, x \mapsto \left\{\begin{array}{ll} 3x^2 &x>0\\ 0 & x=0\\ -3x^2 & x <0 \end{array}\right.$$