Let $f:\mathbb R \rightarrow \mathbb R$ be a continuous function and $x_0 \in \mathbb R$ such that f is differentiable on both intervals $(-\infty, x_0]$ and $[x_0, +\infty)$. Prove or disprove that there exist two functions $g, h : \mathbb R \rightarrow \mathbb R$ differentiable everywhere such that

$$ f(x) = g(x) + h(x)|x - x_0|\ \ \forall x \in \mathbb R. $$

This feels like it characterizes every non-differentiable point of a continuous function in terms of absolute values but I couldn't come up with a function to disprove nor I was able to construct $g$ and $h$.

Help and directions appreciated.

  • $\begingroup$ Note that there might be non-differentiable points where the one-sided limits do not exist, like in $\sqrt{|x|}$, or where the non-differentiable points accumulate somewhere. In both cases you won't have differentiability on $(x_0-\varepsilon,x_0]$ and $[x_0,x_0+\varepsilon)$, so it certainly doesn't characterize every non-differentiable point of a continuous function. $\endgroup$
    – Christoph
    Dec 8, 2018 at 7:32

2 Answers 2


Here is a hint:

Suppose $\phi(x) = \begin{cases} ax, & x < 0 \\ bx, & x \ge 0 \end{cases}$, note that we can write $\phi(x) = {b-a \over 2} |x| + {a+b \over 2} x$.

  • 1
    $\begingroup$ This is a "shining" hint!! (+1) $\endgroup$
    – Robert Z
    Dec 8, 2018 at 8:04
  • $\begingroup$ copper.hat. As always, very nice! +) $\endgroup$ Dec 8, 2018 at 8:41

Hint (to be read after copper.hat hint).

Let us consider the following two differentiable extensions of $f$: $$F_+(x) = \begin{cases} f(x), & x \ge x_0, \\ f(x_0)+f'_+(x_0)(x-x_0), & x \le x_0, \end{cases}$$ and $$F_-(x) = \begin{cases} f(x), & x \le x_0, \\ f(x_0)+f'_-(x_0)(x-x_0), & x \ge x_0. \end{cases}$$ Then $F:=F_+ + F_-$ is differentiable in $\mathbb{R}$ and $$F(x)-f(x)=f(x_0)+\begin{cases} f'_+(x_0)(x-x_0), & x \le x_0, \\ f'_-(x_0)(x-x_0), & x \ge x_0, \end{cases}$$ that is $$F(x)-f(x)=f(x_0)+f'_+(x_0)\cdot \frac{x-x_0 -|x-x_0|}{2} +f'_-(x_0)\cdot \frac{x-x_0 +|x-x_0|}{2}.$$ Can you take it from here?

  • $\begingroup$ Thanks, got it. $\endgroup$
    – halley
    Dec 8, 2018 at 8:24

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