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I'm having trouble with this problem from an old analysis qual: Find a function $f$ such that for $p\in (1,\infty)$, $f$ is in $L^p(\mathbb{R})$ only when $p=4$.

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marked as duplicate by user147263, JimmyK4542, Clayton, user98602, Aditya Hase Dec 14 '14 at 2:29

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  • $\begingroup$ Can you find a function only in $L^p(\Bbb R)$ for $p\ge 4$? $\endgroup$ – Quang Hoang Dec 13 '14 at 22:47
  • $\begingroup$ I'm having trouble there too. I think once I get that the rest shouldn't be too hard. I can do p>4, but not sure about greater or equal to 4. Maybe use log somehow? $\endgroup$ – algebro Dec 13 '14 at 22:59
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    $\begingroup$ Related (but not the same): math.stackexchange.com/questions/1040873/… $\endgroup$ – PhoemueX Dec 13 '14 at 23:40
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Hint: When is $$ f(x) = \frac{1}{|x|^\alpha} $$ $p$-integrable on $(-1,1)$? On $\mathbb{R}-(-1,1)$?

General advice: To control a function's integral norms you generally need to control two things: variance (e.g. oscillation, singularities) and spread (e.g. decay). Higher $p$-norms put more weight on variance and less weight on spread; this trade-off is always present. One way to find functions in a specified $L^p$ is to determine the highest variance and the lowest spread the $p$-norm allows. For a function that exhibits exactly this variance and spread, moving upwards in $p$ weights variance too far, and moving downwards in $p$ weights spread too far. This is one way to think of the hint: pick two values of $\alpha$ to cherry-pick the variance of $f$ at the singularity and the spread of $f$ away from it.

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