Show that there does not exist any function $\phi$ such that $\phi'(x)=f(x)$ for given $f(x)$

Question

1. $f(x)= [x], x\in[0,2]$, where $[x]$ is the greatest integer function.

My attempt:

If possible, let such a $\phi$ exist. Then,

$\phi'(x)=0, x\in[0,1)$

$=1, x\in[1,2)$

$=2, x=2$

So, $\phi'(0)=[0]=0 \neq \phi'(2)=[2]=2$ and $\phi$ is differentiable on $[0,2]$

So, by Darboux Theorem, $\phi'$ must assume every real value between $0$ and $2$, which is a contradiction. Hence, such a $\phi$ does not exist.

Is this proof correct? If so, the second part creates a problem for me.

2. $f(x)=x-[x], x\in[0,2]$

Here both $\phi'(0)=0$ and $\phi'(2)=0$, so how can I apply Darboux Theorem?

Answer 1

In both cases, observe that the given $f$ has a simple discontinuity at $1$, so it does not satisfy the intermediate value property and therefore cannot be a derivative by Darboux's theorem.

Answer 2

Suppose $g$ is an antiderivative for $f$. Then on the in interval $(0, 1)$, it's constant; on the interval $(1, 2)$ it looks like $x$ (plus a constant). Let's go ahead and assume the constant is $0$ on the first interval.

Now $g$ is supposed to be differentiable, so it's gotta be continuous. So far we have \begin{align} g(x) = \begin{cases} 0 & 0 \le x < 1 \\ x + a & 1 < x < 2 \\ 2 & x = 2 \end{cases} \end{align} To make $g$ continuous, we need $g(1) = \lim_{x->1^{-}} g(x) = 0$, so we need $a = -1$. So now we have \begin{align} g(x) = \begin{cases} 0 & 0 \le x \le 1 \\ x -1 & 1 \le x < 2 \\ 2 & x = 2 \end{cases} \end{align}

But this function is not continuous at $2$. So the only possible antiderivative (up to an additive constant) turns out not to be continuous, hence cannot be an antiderivative.