# Find a function in $L^p(\mathbb{R})$ only for $p=4$ [duplicate]

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 $p=4$.

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

• Can you find a function only in $L^p(\Bbb R)$ for $p\ge 4$? – Quang Hoang Dec 13 '14 at 22:47
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.