Does dividing a function by a smooth function result in a smooth function?

[tronter]

If $$f \cdot f$$ and $$f \cdot f \cdot f$$ is smooth, does it follow that $$f$$ is smooth?

So does $$f \cdot f \in C^{\infty} \ \text{and} \ f \cdot f \cdot f \in C^{\infty} \Rightarrow f \in C^{\infty}$$?

Maybe we could generalize a bit more: Given that $$f^{n} , f^{n-1} \in C^{\infty}$$ does it follow that $$f \in C^{\infty}$$ (where $$f^n$$ is the function raised to some power $$n$$)?

 

[Hurkyl]

Well, what does dividing tell you?

 

[tacman]

First, it is important to clarify what is meant by a "smooth function". In mathematics, a smooth function is one that has derivatives of all orders, meaning it is infinitely differentiable. This is denoted by the notation $C^{\infty}$.

Now, to answer the question, dividing a function by a smooth function does not necessarily result in a smooth function. This can be seen by considering the function $f(x) = \frac{1}{x}$. This function is not smooth at $x=0$ because it is not defined there. However, if we divide it by a smooth function, say $g(x) = x^2$, the resulting function $h(x) = \frac{1}{x^3}$ is still not smooth at $x=0$.

In general, dividing a function by a smooth function may result in a function that is smoother than the original, but it does not guarantee that it will be smooth.

As for the second part of the question, if $f^n$ and $f^{n-1}$ are both smooth, it does not necessarily follow that $f$ is also smooth. This can be seen by considering the function $f(x) = |x|$. This function is smooth for all $x \neq 0$, but it is not differentiable at $x=0$.

In conclusion, dividing a function by a smooth function does not always result in a smooth function, and having a certain number of smooth functions raised to different powers does not guarantee that the original function is also smooth. It ultimately depends on the specific functions involved and their properties.