If $\varphi \in C^{0}(\bar{\Omega}) \cap C^{2}(\bar{\Omega} \setminus \left\{0\right\})$, does it imply that $\varphi \in L^{2}(\Omega)$?
1 Answer
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If $\Omega$ is bounded, then yes. If $\Omega$ is unbounded, then no. If $\Omega$ is bounded and $\varphi\in C^0(\bar{\Omega})$, then $\varphi$ is bounded on $\bar{\Omega}$ and hence $\varphi\in L^2(\Omega)$. If $\Omega$ is any unbounded domain, then you can find even a $C^\infty$ smooth radial function $\varphi(x)=f(|x|)$ that growths so fast that $\varphi\not\in L^2(\Omega)$.
answered Apr 15, 2018 at 2:59
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$\begingroup$ What if $\Delta \varphi =0 $ in $\Omega$? $\endgroup$ Commented Apr 15, 2018 at 3:00
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$\begingroup$ I think I figured the answer. Thanks piotr $\endgroup$ Commented Apr 15, 2018 at 3:05
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$\begingroup$ in fact, If $\Omega$ is bounded, I just need $\varphi \in C^{0}(\bar{\Omega})$ to have $\varphi \in L^{2}(\Omega)$, correct? $\endgroup$ Commented Apr 15, 2018 at 3:08
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$\begingroup$ I modified my answer so it should solve your problem. $\endgroup$ Commented Apr 15, 2018 at 5:16
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3$\begingroup$ I think it would be more appropriate to ask this question on Math.StackExchange. Mathoverflow is for research level problems and this one is not. $\endgroup$ Commented Apr 15, 2018 at 12:16