Ngô Quốc Anh

November 6, 2008

Differentiability via weird argument

Filed under: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 2 — Ngô Quốc Anh @ 16:37

Question. Suppose that $f:\mathbb R \to \mathbb R$  is such that $f$ is continuous and $|f|$ is differentiable. Should $f$ also be differentiable?

Proof. Let $g=|f|$. Since $y=f(x)$ satisfies the differentiable equation $y^2=g(x)^2$, $f$ is differentiable at all points with $f(x) \neq 0$ by the implicit function theorem and $2yy'=2gg'$,

and $\displaystyle f'(x)=\frac{g(x)g'(x)}{f(x)}$.

Where $f(x)=0$, we note that $g$ has a minimum and $g'(x)=0$. This gives $\displaystyle\lim_{t\to 0}\left|\frac{f(x+t)}{t}\right|=0$

and $f$ is differentiable by the definition.

Comment. The only points of interests are the zeros of $f$ since $f$ has the same sign in some neighborhood of points which are not roots. So there goes half the work. Geometrically it doesn’t make sense for $|f|$ to have anything but derivative 0 at roots of $f$ and this can be verified by taking left hand and right hand limits of the quotient.