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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?

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  • $\begingroup$ 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. $\endgroup$
    – Lubin
    Jan 22 at 19:06
  • $\begingroup$ 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 $x$. $\endgroup$
    – WimC
    Jan 22 at 19:07
  • $\begingroup$ 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. $\endgroup$
    – zwim
    Jan 22 at 19:21
  • $\begingroup$ @Lubin: Your thought is correct. $\endgroup$
    – Bernard
    Jan 22 at 20:19
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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.

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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 $-1$ ($1$).

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"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 $f'(a)<\lambda<f'(b)$, 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 $\lambda=\frac12$.

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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 $g'(x) = \operatorname{sgn}(x)$.

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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 $0$.

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