# Covering up discontinuities to create analyticity

The floor function, $\lfloor x \rfloor$ , has a "jump" at the integers where its derivative ceases to exist. Everywhere else, its derivative is zero. Now, I wish to multiply the floor function by another function, preferably analytic, such that the product becomes an analytic function. My idea was that since by the product rule the derivative of $ab$ is given by: $$(ab)' = ab'+a'b$$ if the function I multiply the floor function by has zeroes at the integers, it could make the discontinuity of the derivative of the floor function disappear by multiplying it by 0. An obvious start would be $\sin(\pi x)$, but I have already tested this and the function $\sin (\pi x) \lfloor x\rfloor$ is indeed continuous, but its derivative is discontinuous and jumps at the integers. Hence, continuity is easy. But analycity?

Anyone have any ideas?

Edit: I have graphed $\sin(\pi x)^2 \lfloor x\rfloor$, and it seems to be continuous and have a continuous derivative. Can anyone prove or disprove that it is analytic or give a hint how one could go about doing that?

Edit 2: No, that isn't analytic either. I get the feeling multiplying by higher and higher order zeroes(at the integers) allows more an more derivatives to be continuous, but eventually a higher order derivative fails. I would need an infinite order zero at the integers function. Is there such a thing?

Edit 3: I stand corrected, I forgot to mention that since the floor function is clearly 0 from 0 to 1, the product cannot be analytic unless it's zero. I was only aiming for analycity outside the 0 to 1 range.

Edit 4: Daniel Fischer has proven that what I am attempting is impossible. The question is closed. Thank you.

• A function with second-degree zeros at the integers might help, such as $\sin^2(\pi x)$. – Arthur Aug 1 '14 at 14:16
• Also, curiosity: why multiply when you could convolve? Do you have an especially compelling reason to want it to be multiplication? – Adam Hughes Aug 1 '14 at 14:20
• If you want analytic, $0$ is the only way, since for analytic functions we have the identity theorem, saying that if two analytic functions agree on a set having an accumulation point in their domain, they agree on their entire domain, provided that is connected. So if $f$ is an analytic function such that $x\mapsto f(x)\cdot \lfloor x\rfloor$ is analytic, then since $f(x)\lfloor x\rfloor \equiv 0$ on $(0,1)$, it follows that $f(x)\lfloor x\rfloor \equiv 0$ on all of $\mathbb{R}$. – Daniel Fischer Aug 1 '14 at 15:12
• @Semiclassical It's not real-analytic. All derivatives at $0$ vanish, but the function is not identically $0$. – Daniel Fischer Aug 1 '14 at 15:17
• @Assaultous2 That doesn't change anything. We have $f(x)\cdot \lfloor x\rfloor \equiv f(x)$ on $(1,2)$, hence by analyticity $f(x)\lfloor x\rfloor \equiv f(x)$ everywhere, hence $f(x) = 2f(x)$ on $(2,3)$, so $f\equiv 0$ on $(2,3)$, so $f\equiv 0$. – Daniel Fischer Aug 1 '14 at 15:19