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My math problem involves using a theorem that requires $f(x)=(x-6)^2$ to be in $C_0^2$. I'm trying to understand what $C_0^2$ means and how to check whether a function belongs to it. The course I'm taking is focusing on something very different and the book assumes that we have prior knowledge of this.

EDIT: x is in R

EDIT: I'm trying to use the following theorem where $f(x)$ needs to be in $C_0^2$ enter image description here

dark blue
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  • I believe Feynman-Kac does not require the function to be compact. Often one proves something for compact functions and then generalizes it by an approximation argument. See, for instance, these lecture notes, which include no requirement that $f$ be of compact support. – mboratko Jun 29 '14 at 13:49
  • I meant to include: Your book may prove only the compact case (perhaps this is a simplifying assumption in the proof) or perhaps your book later mentions that compactness is not needed. The wikipedia entry for the Feynman-Kac formula also does not seem to include the compactness condition for the initial condition, at least explicitly, but often for applied PDEs compactness is assumed implicitly. An inspection of the proof should indicate if it's really needed. – mboratko Jun 29 '14 at 13:55

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A function in $C_0^2$ is twice continuously differentiable and has compact support. If you have an explicit function $f(x)$, you can check if it is in $C_0^2$ by differentiating it twice and making sure that $f''(x)$ is also continuous. Furthermore, you need to be sure that your function is nonzero only on a compact set.

In your example, the function is a polynomial, so it is infinitely differentiable and therefore trivially in $C^2$, however unless the domain is restricted it does not have compact support (the values for which it is nonzero are unbounded).

Furthermore, one cannot arbitrarily restrict the domain and expect the function to continue to be differentiable, indeed if the domain is restricted at points where $f(x)$ is nonzero the function will no longer even be continuous. A common argument used when compact support is needed is to multiply the function by an appropriate bump function.

mboratko
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  • thank you for your help, x is in R, does it provide the compact support? – dark blue Jun 29 '14 at 13:22
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    No, in $\mathbb R$ the only compact sets are closed and bounded intervals, i.e. sets of the form $[a,b]$. You should check how the compactness is used in the theorem you are attempting to use, and see if it is really necessary for your purposes. A standard argument when compactness is needed is to multiply the function by a bump function. – mboratko Jun 29 '14 at 13:25
  • thank you once again, this very helpful! I have updated the original question with the theorem I'm trying to use so it provides more background.. – dark blue Jun 29 '14 at 13:38