# Symmetric derivative of nondifferentiable function $f$ must be zero?

Let $$f$$ be a continuous function.

We know $$f$$ is differentiable at $$x = a$$ if the limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ exists. (Normal derivative)

And, if $$f$$ is differentiable at $$x = a$$ then we can write $$f'(a)$$ as $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}$$

But the limit (symmetric derivative) $$\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$$ can exist when $$f$$ is not differntiable at $$x = a$$

For example, a function $$x \to |x|$$ is not differentiable at $$x = 0$$ but the limit exists : $$\lim_{h\to 0}\frac{f(0+h)-f(0-h)}{2h} = \lim_{h\to 0}\frac{|h|-|-h|}{2h} = 0$$

I checked some examples ($$f$$ is not differntiable but the symmetric derivative exists) and I got zero for all $$f$$.

So my question is :

1. If $$f$$ is not differentiable at $$x=a$$ and symmetric derivative exists at $$x=a$$, then the symmetric derivative of function $$f$$ at $$x=a$$ must be zero? (How to prove? or I want to know counterexample.)

2. Is there any nice approach to understand symmetric derivative? (I understood this as a slope of two points.)

• What about $f(x)=x$ if $x>0$ and $f(x)=-2x$ if $x\le 0$? Oct 9, 2021 at 8:12
• yes, any function $f:x\mapsto g(x)+c|x|$ with $g$ differentiable and $c\not =0$ produces a counterexample. Oct 9, 2021 at 8:28
• @Giulio: I think you want to impose $g'(0) \neq 0$ as well. Oct 9, 2021 at 11:37
The symmetric derivative is basically the arithmetic mean of the left-hand and right-hand slopes. So if they're equal in magnitude but of opposite signs, you will get $$0.$$ For instance, if we're evaluating the derivative at $$a=0$$, then any even function such as $$y=|x|$$ will yield a symmetric derivative of $$0.$$
However, any piece-wise (or non-differentiable) function with "uneven" slopes will have a non-zero symmetric derivative. A simple example is $$f(x)=x \text{ for }x>0$$ and $$f(x)=-2x \text{ for } x≤0$$ as pointed out by @stephenkk in the comments.