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Does there exist a function f such that $f'(x)=\lfloor x \rfloor$ for all $x$ in $\mathbb{R}$?

I was thinking no, since if you draw it out and look at the integral, it should be an imaginary number, right?

But how do I explain it?

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No. You could appeal to Darboux's Theorem. – David Mitra Dec 4 '12 at 19:55
I was actually thinking about it, also. Can i use this as a counter example? consider $1/2\in[0,4]$, by Darboux, then there exists $x\in(0,4)$ such that $f'(x)=1/2$ – Akaichan Dec 5 '12 at 1:07
Yes, that would show that no such $f$ exists as in your post (it's not really a counterexample, though; rather, part of a proof that there is no such $f$ via contradiction). – David Mitra Dec 5 '12 at 1:28
up vote 3 down vote accepted

Well, no if you wish for $f$ to be differentiable (as pointed out by David Mitra, Darboux's theorem can be used to formalize this). However, you could consider weak derivatives. Then $f$ could be a picewise linear function, the step of which is $\lfloor x \rfloor$, e.g. (there are many because of the integration constant):

$\hspace{40pt}$enter image description here

Be aware that this is not regular antiderivative, on the other hand it's quite a common occurence in practical applications. The shape is in fact almost the same as the example of @icurays1, still it might be just more intuitive to make it continous. I hope I didn't confused you ;-)

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The integral certainly won't be an imaginary number, but you'll have an issue at the integers because the floor function has a jump discontinuity there, which will manifest itself as a "kink" in the integral. The function below will have $f^\prime=\lfloor x\rfloor$ at all noninteger points:

$$ f(x)=nx,\; x\in [n,n+1)\quad n\in\Bbb{Z} $$

However, this function is not differentiable if $x=k$ where $k\in\Bbb{Z}$. As suggested by the comment, it can be shown that in fact no such function can exist that has $f^\prime(x)=\lfloor x\rfloor$ for all $x$.

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