# No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$

How can I prove for certain that there exists no differentiable function $$g(x)\colon [-1,1]\rightarrow \mathbb{R}$$ with

$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1, & \text{for x > 0} \\ 0, & \text{for x=0} \\ -1, & \text{for x<0} \end{cases}$$

I'm currently jumping to the absolute value function but that's not differentiable at $$x = 0$$ and that doesn't exactly disprove the existence of others.

Looking at the limits \begin{align} \lim_{h\to 0^{+}} g'(h) &=1\\ &= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\ &\neq \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\ &= -1\\ &= \lim_{h\to 0^{-}}g'(h)\\ &\neq 0\\ &= g'(0) \end{align} which implies discontinuity at $$h = 0$$.

Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?

• I think it’s the case that the derivative of a differentiable function, though not necessarily continuous, must satisfy the intermediate value property. If this is right, then since the sgn function does not have IVP, you have a proof. Jan 22 at 19:06
• Derivative zero at $x=0$ means that the difference quotient is close to zero for $x$ near $0$ and then the mean value theorem shows that this difference quotient equals the derivative at some other point strictly between $0$ and $x$.
– WimC
Jan 22 at 19:07
• You can use $f_n(x)=\frac 1n\ln(\cosh(nx))$ as an approximation of $|x|$ with a flat at $x=0$ as $\tanh(nx)$ is itslef an approximation of signum function.
– zwim
Jan 22 at 19:21
• @Lubin: Your thought is correct. Jan 22 at 20:19

Hint Use Darboux Theorem. If you are not familiar with it, mimic and simplify (due the fact that you have a very simple function) one of its proofs.

If such a $$g$$ existed, we could take without loss of generality $$g(0)=0$$. Then $$g(x)=|x|$$, but this achieves a contradiction: $$g^\prime(0)$$, rather than being $$0$$, does not exist because the left- (right-)derivative is $$-1$$ ($$1$$).

"Can I use the contraposition from Darboux's Theorem for this specific case?"

Yes. Darboux's theorem says that if $$f:[a,b]\to\mathbb{R}$$ is a differentiable function and suppose $$f'(a)<\lambda, then there is a point $$x\in(a,b)$$ such that $$f'(x)=\lambda$$.

The contraposition says that if $$f'$$ does not satisfy the "intermediate property above", then $$f$$ cannot be differentiable on $$[a,b]$$.

The existence of $$g$$ in your question would contradict this theorem: you may consider for instance $$\lambda=\frac12$$.

Suppose that such a function $$g$$ existed. Then based on your definition of $$g'$$ we can conclude that $$g(x) = \begin{cases} -x+c_1, &x<0\\ a, &x = 0\\ x+c_2, &x > 0 \end{cases}$$ where we don't know the value of $$a$$. The only question is whether or not there are some choice of constants to make $$g'(x) = \operatorname{sgn}(x)$$. If that were to work, then following would work out: $$0 = g'(0) = \lim_{h\to 0}\frac{g(h) - g(0)}{h}.$$ Because the definition of $$g$$ differs between the left and right of $$0$$, we need to consider the left and right limits, though as we will see it suffices to consider the right-hand limit: $$\lim_{h\to 0^{+}}\frac{g(h) - g(0)}{h} = \lim_{h\to 0}\frac{h+c_2 - c_2}{h} = \lim_{h\to 0}\frac{h}{h} = 1 \ne 0.$$ Thus, we can conclude that there is no such function $$g$$ with $$g'(x) = \operatorname{sgn}(x)$$.

Assumed a g(x) exists.

1)Solution for $$x>0:$$ $$g(x)=x +a;$$

2)Solution for $$x <0:$$ $$g(x)=-x+b;$$

Since differentiability implies continuity: $$a=b=0;$$

$$3)\lim_{x \rightarrow \pm 0} \dfrac{g(x)-0}{x}$$ does not exist,

hence not differentiable at $$0$$.