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Is the following true? with proof. In other words check whether $\partial^{2}_{r}$ commutes with $\partial^{-2}_{r}$. $$\partial^{-2}_{r}\partial^{2}_{r}=\partial^{2}_{r}\partial^{-2}_{r}$$ Side Note: In case of $\Box^{-1}\Box\neq \Box\Box^{-1}$.

Wiliam
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Hint
Check out the most basic case: One variable and the space $\mathbb R[x]$. Is $$\frac{\mathrm d^2}{\mathrm dx^2} \int_0^x \int_0^s f(t) \mathrm dt \ \mathrm ds = \int_0^x \int_0^s \frac{\mathrm d^2}{\mathrm dt^2} f(t) \mathrm dt \mathrm ds \qquad?$$ This question is answered here (not really a duplicate, but will answer your question as well)

Community
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AlexR
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  • thanks AlexR, they indeed are not equal. the RHS in zero while the LHS is non zero! – Wiliam Feb 01 '15 at 16:12
  • @Ilia can be zero... If you have $\deg f > 2$ and the quadratic, linear and constant terms zero, the equality actually works ^^ – AlexR Feb 01 '15 at 16:17
  • yet how do you define the same integral for the particular case above? is this also true for above? – Wiliam Feb 01 '15 at 16:17
  • @Ilia Yes, it is. If the 1D case fails already you can pick $\phi(r,\theta) = f(r)$ with a counter-example function $f$ to pull it up into higher dimensions. – AlexR Feb 01 '15 at 16:18
  • Great thanks :) – Wiliam Feb 01 '15 at 16:20
  • Yes, it works rarely, let alone in general. After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. – AlexR Feb 01 '15 at 21:09
  • Just did :) a bit new to this website... – Wiliam Feb 01 '15 at 21:54
  • No problem :) We all were at some point. The help center is a good place to start exploring. – AlexR Feb 01 '15 at 21:56
  • Now another question I have in mind is what is happening in the following cases? $$[\Box,\frac{1}{\Box}]\mathcal{O}=0 ? $$ $$[\nabla,\frac{1}{\nabla}]\mathcal{O}=0 ? $$ $$[\nabla^2,\frac{1}{\nabla^2}]\mathcal{O}=0 ? $$ $$[\partial^{2}{r},\frac{1}{\partial^{2}{r}}]\mathcal{O}=0 ?$$ in the case of $\Box$ we know they do not commute.. and $\mathcal{O}$ is operator. – Wiliam Feb 01 '15 at 22:29
  • You should ask that as a new question, showing also how you came to this and what your thoughts are to ensure the best possible answers- – AlexR Feb 01 '15 at 22:32