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I'm reading chapter 8 of the pde book written by EVANS. It's about variation method of pde. In page 433. what's the mean of $C_c^\infty(U)$? why the test function $v$ is in this space?

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thank you!

hongjinwu
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  • The space $C^\infty_c(U)$ is the space of compactly supported smooth functions on $U$, if this is what you are asking about. – Malkoun Jun 05 '18 at 10:30
  • thanks for your answer. I have calculate as follows: $\int_U (-V(L_{p_i}){x_i}+L_z V)dx+\sum L{p_i} V|\Omega =0$ ,so $\sum L{p_i} V|_\Omega =0$ needs to be zero. But why $v$ belongs to $C_c^\infty(U)$? – hongjinwu Jun 05 '18 at 11:46
  • I am not on my laptop now. I can reply later. – Malkoun Jun 05 '18 at 11:48
  • look forward your answer :) – hongjinwu Jun 05 '18 at 11:49
  • You basically want the boundary integral to vanish, when you do an integration by parts, which is why you require $v$ to have compact support. Does that answer your question? – Malkoun Jun 05 '18 at 15:27
  • I don't really understand why $\sum L_{p_i}V|_{\Omega}=0$ means $v$ have compact support. – hongjinwu Jun 06 '18 at 07:37
  • I don't understand. I only see the picture that you have linked to (I don't have Evans' book near me). Are you sure that Evans is deducing that $v$ has compact support? Or is he assuming it? Remember that $v$ is the test function. I have seen arguments like that, and usually one assumes that the test function is of class $C^\infty_c$, and then makes an integration by parts. I hope my comments help a little. – Malkoun Jun 06 '18 at 10:26
  • I understand it ! thank you very much! – hongjinwu Jun 08 '18 at 12:08

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$C_{c}^{\infty}(U)$ denote the space of infinitely differentiable functions with compact support in $U$. By definition, the function belonging to this space called a test function. See page 242 of Evans's book for more details.

S. MAATOUK
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