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Let $ f $ be in $ L^1(\Omega) $ where $ \Omega $ is an open subset of $ \mathbb{R}^n $. Also, assume that $ B(x_i,r_i) $ is a collection of disjoint open balls in $ \Omega $ such that $ B(x_i,2r_i) \subseteq \Omega $. I was wondering if there is a way that we can find a function $ g \in L^1(\Omega) $ and possibly a constant $ C $ such that

$ \int_{B(x_i,2r_i)} | f | \le C\int_{B(x_i,r_i)} | g | $, for all $ i \in \mathbb{N} $

(If $ f \in L^p(\Omega) $ with $ p > 1 $, then I can find such $ g $ via the Hardy-Littlewood maximal function.)

Thank you so much.

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No chance. The twice larger balls can easily have a common point, so if you put a Dirac point mass there, you'll blow up the $L^1$-norm of $g$. –  fedja Mar 15 '13 at 20:34

3 Answers 3

No. Take $\Omega = \mathbf{R}$, $$f(x) = \left\\{\begin{array}{cc} x^{-1/2} & |x| < 1\\\\ 0 & |x| \geq 1 \end{array}\right.$$

$x_i = 3^i$, and $r_i = x_i/2$ $$\int_{B(x_i,2r_i)} |f| = \int_0^{1} x^{-1/2} dx = 2$$

Assume there exists a $g$ in $L^1(\mathbf{R})$ $$\int_{\Omega} |g| \geq \sum_i \int_{B_i} |g| \geq C^{-1} \sum_i \int_{B(x_i,2r_i)} |f| \geq C^{-1} \sum_i 2 = \infty$$ which contradicts the hypothesis.

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The answer is negative. Let $n=1$, $\Omega=(0,\infty)$ and $f(x)=\min\bigl(1/x^2,1\bigr)$. Let $x_i=2\cdot 4^i$ and $r_i=4^i$. Then $B(x_i,2r_i)$ contains $(0,1)$, and so $\int_{B(x_i,r_i)} |f|\geq \int_0^1 |f|=1$. Therefore, the putative function $g$ satisfies $$\int_0^\infty |g|\geq \sum_{i=1}^\infty \int_{B(x_i,r_i)} |g|\geq \sum_{i=1}^\infty \frac{1}{C}\int_{B(x_i,r_i)} |f|\geq \sum_{i=1}^\infty \frac{1}{C}=\infty.$$

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Since the $B(x_i, r_i)$ are disjoint, you can take $g(x) = f(x_i+2(x-x_i))$ for $x \in B(x_i, r_i)$ and $g(x)=0$ otherwise.

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