Does there exist $f: \mathbb R \rightarrow \mathbb R$ with $f'(x) = h(x)$ for all $x\in\mathbb R$.

Question

Let $h: \mathbb R \rightarrow \mathbb R$ be given by $h(x) = -1$ if $x\le0$ and $h(x) = +1$ if $x\gt0$ (a Heaviside step variant). Does there exist $f: \mathbb R \rightarrow \mathbb R$ with $f'(x) = h(x)$ for all $x\in\mathbb R$.

My first impression allows $f$ to take the form of $f(x) = |x|$. It's not exactly the same because absolute is not differentiable at $x=0$ but it raises an intuition that such a function does not exist.

Then I begin to think how $|x|$ is discontinuous at $x=0$ and I wonder if I should apply a analogous proof here i,e. I suppose such an $f$ does exist such that $f'(x) = h(x)$ for all $x\in\mathbb R$, and highlight this inconsistency between the left and right limit at 0. The only problem is, I don't know how to properly approach this. Any pointers?

This follows from Darboux, but I'd like to see someone give a more elementary argument. — Randall, Oct 09 '18 at 01:56
You could also solve the differential equation at points other than $x = 0$, obtaining a unique solution (up to a constant) each for the positive and negative $x$ values. You can then use the continuity of your anti-derivative to conclude how the two constants must be related, and the value of the anti-derivative at $x = 0$. This function will not be differentiable at $x = 0$. — Theo Bendit, Oct 09 '18 at 02:00
By mean value theorem a derivative can't have jump discontinuity like your heavy side step function has. — Paramanand Singh, Oct 09 '18 at 04:39
See first part of the answer https://math.stackexchange.com/a/577978/72031 — Paramanand Singh, Oct 09 '18 at 04:42
@TomRyddle . I edited the Q. There was an obvious (to me) typo in the def'n of $h$. — DanielWainfleet, Oct 09 '18 at 04:46

score 1 · Answer 1 · answered Oct 09 '18 at 05:21

Suppose $f'(y)=-1$ for $y<0$ and $f'(x)=+1$ for $x>0$, and that $f(z)$ is continuous at $z=0.$

By the Fundamental Theorem of Calculus, if $y<0$ then $f(y)=f(-1)+\int_{-1}^y f'(z)dz=f(-1)-y-1,$ while if $x>0$ then $f(x)=f(1)+\int_1^xf'(z)dz=f(1)+x-1.$

So $f(-1)-1=\lim_{y\to 0^-}f(y)=f(0)=\lim_{x\to 0^+}f(x)=f(1)-1.$

Therefore $f(z)=f(0)+|z|$ for all $z$, so $f(z)$ is not differentiable at $z=0.$

score 1 · Answer 2 · answered Oct 09 '18 at 05:31

(Please if someone could check my reasoning)

As plenty have mentioned, this can be proven via the Darboux theorem. (Recap) Given a function $f:[a,b]\rightarrow \mathbb R$ that is differentiable. Suppose $y\in\mathbb R$ such that $f'(a) \lt y \lt f'(b)$ or $f'(a) \gt y \gt f'(b)$ then there exists a $c \in (a,b)$ such that $f'(c) = y$. (Should remind you of the intermediate value theorem for continuous functions).

So in this case we build a contradiction. Suppose there exists such a function, indeed call it $f$, that is differentiable and the derivative is given via our defined function $h$. Well...

If $f$ is differentiable on $\mathbb R$, it must also be differentiable on the interval $[-1,1]$ and since $0 \in \mathbb R$ and $f'(-1) = -1 \lt 0 \lt 1 = f'(1)$, there exists some $y\in(-1,1)$ such that $f'(y) = 0$. Yet clearly by definition of $h$, this is impossible.

Conclusion: There exists no function, $f$, with such property.

score 0 · Answer 3 · answered Oct 09 '18 at 05:13

0

If f'(x) is the Heaviside step function, the solution to the problem is it's integral, which is the ramp function R(x) = xH(x).

answered Oct 09 '18 at 05:13

John

102

Does there exist $f: \mathbb R \rightarrow \mathbb R$ with $f'(x) = h(x)$ for all $x\in\mathbb R$.

3 Answers3