# Showing that a function doesn't exist

Does there exists a differentiable function $f:\mathbb{R} \to \mathbb{R}$ such that:

1. $f(0) = 0$.

2. $f'(x) = 1$ for all $x \in \mathbb{R}$ with $f(x) \lt 0$.

3. $f'(x) = -1$ for all $x \in \mathbb{R}$ with $f(x) \ge 0$.

Clearly such function doesn't exist because it is monotone decreasing at 0, but increasing for $x > 0$ which is impossible however I didn't manage to prove this.

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Can you visualize such a function? Can you compute the derivative at $x = 0$ using limits from both sides? – shardulc Mar 27 at 16:01
The open set $\{x : f(x) < 0\}$ cannot contain an interval of the form $(a,\infty)$ (why?). Thus, it has a boundary point in $\mathbb R$. Consider such a boundary point. – Friedrich Philipp Mar 27 at 16:19
The conditions mean that $f(x) = -|x|$ for all $x$, which is not differentiable at $0$. However, to prove that $|x|$ is not differentiable at $0$ requires a method like those already offered. – Paul Sinclair Mar 27 at 22:38

First solution:

Notice $f'(0)=-1$, take $\epsilon=1/2$, then there is $\delta>0$ so that if $x\in (-\delta,\delta)$ then $\frac{f(x)-f(0)}{x-0}\in (-3/2,-1/2)$. In particular $f(x)$ has a different sign than $x$, so take a fixed $x_0$ in $(0,\delta)$, then $f(x_0)<0$.

We have $f'(0)=-1$ and $f'(x_0)=1$. We conclude with Darboux's theorem that there is $x\in (0,x_0)$ with $f'(x_0)=0$, a contradiction.

Second solution:

Because of Darboux's theorem the derivative of a differentiable function cannot take exactly two values.

So the only two options are $f(x)=x+c$ and $f(x)=-x+c$. However, these functions have both positive and negative values, so the derivative should take on two values.

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Please edit your second solution. – Friedrich Philipp Mar 27 at 16:49
what should I change? – Carry on Smiling Mar 27 at 16:51
There's not that much to check, right? – Friedrich Philipp Mar 27 at 17:03
Is that what you meant? Thanks by the way. – Carry on Smiling Mar 27 at 17:04
Yes, that was it. Thank you, too. :o) – Friedrich Philipp Mar 27 at 17:06

Ok, here is an answer that does neither make use of Darboux's theorem nor of the assumption 1. So, assume that only 2. and 3. hold. Suppose that there exists $a\in\mathbb R$ such that $f(x) < 0$ for all $x\in (a,\infty)$. Then $f'(x) = 1$ there which implies that $f(x) = x + c$ for $x\in (a,\infty)$ with some $c\in\mathbb R$. But then $f(x) > 0$ for some $x\in (a,\infty)$, contrary to the assumption that $f(x) < 0$ on $(a,\infty)$. Similarly one shows that $\{x : f(x)\ge 0\}\neq \mathbb R$.

Therefore, there exists some $x_0\in\mathbb R$ such that $f(x) < 0$ for $x < x_0$ and $f(x_0) \ge 0$. Then we have $f(x_0) = 0$ (by continuity of $f$) and $f'(x_0) = -1$. Therefore, $$-1 = \lim_{x\uparrow x_0}\frac{-f(x)}{x_0-x}\ge 0,$$ a contradiction. Hence, a differentiable function satisfying 2. and 3. does not exist.

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Since $f'(0) = -1, f(0) = 0$ it follows that there is a number $\delta > 0$ such that $f(x)$ is positive for all $x \in [-\delta, 0)$ and $f(x)$ is negative for all $x \in (0, \delta]$. This means that $f'(x) = -1$ for all $x \in [-\delta, 0]$ and $f'(x) = 1$ for all $x \in (0, \delta]$. Thus $f'$ has a jump discontinuity at $x = 0$ and this is impossible as derivatives don't have jump discontinuity. Hence no such function $f$ can exist.

The last bit about derivatives not having jump discontinuity can be established via mean value theorem but we can more directly obtain the contradiction for the specific function $f$ of the current question. Let $x \in (0, \delta]$ then we know that $$f(x) = f(x) - f(0) = xf'(\xi)$$ for some $\xi \in (0, x)$. And since $f'(x) = 1$ for $x \in (0, \delta]$ it follows that $f(x) = x$ for all $x \in (0, \delta]$. And this is inconsistent with $f(x)$ being negative for $x \in (0, \delta]$.

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Let's consider $x_1>0$.

If $f(x_1)>0$ by continuity $f(x)>0$ (and therefore $f'(x)=-1$) for any $x \in (x_1-a,x_1+a)$ for some $a$.

Call $x_0=\inf\{t:f(x)>0 \; \forall x \in (t,x_1)\}$ (that we just proved to be non empty).

• If $x_0>-\infty$ by continuity $f(x_0)=0$ but this is impossible because $f'=-1$ in $(x_0,x_1)$ and so $f(x_0)>f(x_1)>0$.

• If $x_0=-\infty$ then $0 \in (x_0,x_1)$ and so $f(0)>0$, that is impossible.

Since $x_1$ is arbitrary we must conclude that $f(x)\leq 0$ and $f'=1$ for any $x \in [0,+\infty)$, but this is still impossible because if $x>0$ we would have $f(x)>f(0)=0$ because $f'=1$ on $[0,+\infty)$ implies $f$ is strictly increasing on $[0,+\infty)$.

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Since $f'(0) = -1$, for some $\delta>0$ we have $f(x) \le - {1 \over 2} x$ for $x \in [0,\delta)$.

In particular, $f(x) <0$ for $x \in [0,\delta)$ and so $f'(x) = 1$.

If $0<y<x < \delta$, we have $f(x)=f(y) +x-y$, and by continuity (letting $y\to 0$) we have $f(x) = x$, which contradicts $f(x) \le - {1 \over 2} x$ for $x \in [0,\delta)$.

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Actually $f'(x)=-\text{sgn}(f(x))$ – Marco Disce Mar 27 at 17:24
@MarcoDisce: Thanks, I completely missed that and was wondering why the other answers were so complicated. – copper.hat Mar 27 at 17:25
@MarcoDisce: I have repaired my answer. – copper.hat Mar 27 at 17:34