The Unit step function(derivative of $y = |x|$) does not take the values from $(0, 1)$. It is said that this function cannot be a derivative of any real valued function. But isn't $y = |x|$ a real valued function, whose domain and range are both real numbers? So, how is this unit step function not a derivative of any real valued function?
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$\begingroup$ Oh, I think you mean the Heaviside Function...but that is not the derivative of the absolute value function, but rather of the ramp function $\;\max\{x,0\}\;$ , very different from $\;|x|\;$ ... $\endgroup$– DonAntonioCommented Mar 21, 2017 at 12:28
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2$\begingroup$ The $| \cdot |$ function is not differentiable at $0$. Although it has a derivative - the $\operatorname{sgn}$ function - in the distributional sense. $\endgroup$– Eman YalpsidCommented Mar 21, 2017 at 12:30
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$\begingroup$ @DonAntonio , I don't know about the heaviside function , see this: pasteboard.co/M5k4qAlYo.png $\endgroup$– BumbleBeeCommented Mar 21, 2017 at 12:41
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$\begingroup$ @Andrew, the |.| function? if f = |x| then |x|/x = f', see: pasteboard.co/M5k4qAlYo.png $\endgroup$– BumbleBeeCommented Mar 21, 2017 at 12:42
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1$\begingroup$ The fundamental theorem of calculus requires the original function (i.e. the derivative) to be continuous. Both the Heaviside step function and the sign function are discontinuous at a particular point $\endgroup$– HenryCommented Mar 21, 2017 at 12:43
1 Answer
$\DeclareMathOperator{\abs}{abs}$By definition, a real-valued function $f$ defined on some non-empty open set $I$ is differentiable on $I$ if $f$ is differentiable at $x$ for every $x$ in $I$.
Let $\abs(x) = |x|$ for $x$ real. The difference quotient at $0$ is $$ \frac{\abs(h) - \abs(0)}{h} = \frac{|h|}{h},\quad h \neq 0. $$ Since the difference quotient has no limit as $h \to 0$ (the one-sided limits exist but are unequal), $\abs'(0)$ does not exist. That is, the absolute value function is not differentiable at $0$.
Consequently, the absolute value function is "not differentiable" on any open interval containing $0$, e.g., is not differentiable on the set of real numbers. The fact that $\abs'(x)$ exists for all $x \neq 0$ is immaterial.
Careful examination of the intermediate value property for derivatives reveals that the domain of $f$ must be an interval. The absolute value function does not contradict the intermediate value property: the function $\abs'(x) = |x|/x$ for $x \neq 0$ fails to satisfy the intermediate value property on a non-empty open real interval $I$ if and only if $0 \in I$, if and only if $\abs$ is not differentiable on $I$.