I have a real-valued function $f$ defined on a ball $B$ of $\mathbb{R}^{N}$, $N\geq2$. I have found a constant $M>0$ such that for all $x\in B$ and $B(x,R)$ (ball of center $x$ and radius $R>0$) such that its closure $\overline{B(x,R)}$ is in $B$, we have $$|f(x)|\leq M\max_{B}|\Delta\phi|,$$ where $\phi$ is a non-negative test function with support in $B(x,R)$ such that $\phi=1$ on $\overline{B(x,r)}$ with $0<r<R$, or on any set containing $\overline{B(x,r)}$ and contained in $B$. Here, $r$ is arbitrary and $ \phi $ is the function given by Urysohn's lemma. My question is: Can I conclude that $f$ is locally bounded on the ball $B$? It seems to me that since the right side of the above inequality depends on $ \phi $, the answer is no. Is it possible to get rid of $ \phi $?
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2$\begingroup$ Is $\phi$ arbitrary? If yes, then you can simply choose $\phi$ in such a way that $|\Delta \phi| \leqslant C_1 (\operatorname{dist}(B^c, B(x, R)))^{-2}$ to get $|f(x)| \leqslant C_2 M (\operatorname{dist}(x, B^c))^{-2}$. $\endgroup$– Mateusz KwaśnickiCommented Sep 17, 2017 at 19:19
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$\begingroup$ $ \phi $ is a non-negative test function with compact support in $B$ that takes the value of 1 on $\overline{B(x,r)}$, or on any other set containing $\overline{B(x,r)}$ and contained in $B$. I didn't get the function $\phi$ you are talking about. Can you please explain? $\endgroup$– M. RahmatCommented Sep 18, 2017 at 0:43
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1$\begingroup$ Again: Is $\phi$ a test function or any test function? (The former meaning "you know exactly which one and we are not allowed to pick another", the latter meaning "it doesn't matter which one, just pick any test function that satisfies the conditions"). If it is the latter, than @MateuszKwaśnicki constructs the function like this: Fix $\phi_0:\mathbb{R}\to\mathbb{R}$ with support [-2,+2] and plateau [-1,+1]. Then use a scaled, rotationally symmetric version of $\phi_0$ as your $\phi$. Scaling with $\lambda$ introduces the factor $\lambda^{-2}$ in 2nd derivatives. Done. $\endgroup$– Johannes HahnCommented Sep 18, 2017 at 1:14
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$\begingroup$ Just pick any test function that satisfies the conditions. Thanks. $\endgroup$– M. RahmatCommented Sep 18, 2017 at 4:42
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$\begingroup$ I have a question: will this $\phi$ be between 0 and 1? $\endgroup$– M. RahmatCommented Sep 20, 2017 at 4:26
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