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I am trying to find such function, that belongs to $L^2([0;1])$ (with standard Lebesgue measure), but not to $L^p([0;1])$, for $p >2$. It is easy to point example when $p \ge 3$, $ p \ge \frac{5}{2}$ or in general $p \in [k; \infty)$, where $k>p$, but I am not sure how to do it for $p \in (2; \infty)$. Thank you in advance for any tips.

MI00
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    @KaviRamaMurthy I may be misunderstanding the question, but the desired function should not depend on $p$ correct? To me it sounds like OP is asking for one function that's not in $L_p$ for every $p > 2$ but is in $L_2$. – jl00 Nov 21 '20 at 05:35
  • @jl00 Yes, exactly. – MI00 Nov 21 '20 at 05:38
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    These may help, seem to be stronger versions of what you are asking: https://math.stackexchange.com/questions/1552821/give-an-example-of-a-function-which-is-in-l2-mathbbr-but-not-in-lp-ma and https://math.stackexchange.com/questions/55170/is-it-possible-for-a-function-to-be-in-lp-for-only-one-p?noredirect=1&lq=1 – jl00 Nov 21 '20 at 05:43
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    Let $E_n$ be disjoint sets with $m(E_n)=\frac 1 {n^{3} ((\ln n)^{2}}$ and $f =\sum_n n\chi_{E_n}$. – Kavi Rama Murthy Nov 21 '20 at 05:46

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