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Jan 30 at 9:09 vote accept Ma Joad
Jan 30 at 2:06 answer added Willie Wong timeline score: 1
Jan 30 at 1:39 comment added Willie Wong @ChristianRemling: the OP means spherical maximal as in "supremum of spherical means". So on $\mathbb{R}$ we are looking at $Mf(x) = \sup_{r > 0} \frac12 \Big( |f(x+r)| + |f(x-r)| \Big)$. If you have a bounded function $f$, we then have $|Mf(x)| \in [\frac12 \|f\|_\infty, \|f\|_\infty ]$ and hence what the OP demands hold. But it fails for (deliberately poorly chosen representatives of) $L^\infty$ functions.
Jan 30 at 1:33 comment added Willie Wong Incidentally, you probably don't want to consider the spherical maximal function anyway on $L^1(\mathbb{R})$, as it isn't even well-defined. (Take a random $L^1(\mathbb{R})$ function $f$, modify it so that $f(n) = |n|$; this new function differs only on a measure zero set from the original, so is still $L^1$. But at every point you find $Mf(x) = +\infty$.)
Jan 30 at 1:26 comment added Willie Wong When $n = 1$ what you are looking at holds essentially by accident. When $n \geq 2$ take $\varphi(x) = f(-x)$ to be the characteristic function of the ball of radius 1 centered at then point $(n,0,0,0,\ldots,0)$. Then even with the "spherical" maximal function this extension of Christian's counterexample works.
Jan 29 at 18:06 comment added Christian Remling The spherical (= symmetric) maximal function is what I had in mind.
Jan 29 at 10:39 comment added Ma Joad @ChristianRemling Thank you. Does spherical maximal function work, though?
Jan 29 at 10:39 history edited Ma Joad CC BY-SA 4.0
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Jan 28 at 23:46 comment added Christian Remling No. If $\varphi(x)=f(-x)=\chi_{(n,n+1)}(x)$, then $(\varphi *f)(0)=\|\varphi\|_1=1$, but $(Mf)(0)\simeq 1/n$.
Jan 28 at 23:26 history edited YCor CC BY-SA 4.0
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Jan 28 at 22:56 history asked Ma Joad CC BY-SA 4.0