# Why can differentiating a function carry it out from Hilbert space?

I was just doing a QM Griffiths Problem. I was able to get it correct, but I have a few questions.

Let $f(x)=x^v$ be defined on $[0,1]$ where $v= 1/2$ then $df/dx = \frac{x^{-1}}2$

Then, we know that the inner product, $\langle f |f \rangle = ln|x|$ on $[0,1]$ diverges at $x=0$. Why, then, does this function cease to exist after differentiating it?

My guess is that it matters less about the function and more about whether its derivative or integral is square-integrable and infinite. In other words, If $df^n/dx^n$ lives in Hilbert space, then you are guaranteed that $f(x)$ also lives in Hilbert space, but not the other way around. Thus, if $f(x)$ is square-integrable and lives in Hilbert space, then any of its integrals will also live in Hilbert space.

• It is in general not true that the primitive (anti-derivative) of a square integrable function is again square integrable. Take $f(x)=1/(1+x{^2})$ and $L{^2}(R)$. – Urgje Oct 14 '15 at 10:24
• A function and its derivatives are not necessarily related when it comes to integrability. There are infinitely many counterexamples of integrable functions whose derivative is not or vice-versa. – yuggib Oct 14 '15 at 10:43