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.

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