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I want to know if $$L^1 [0,1]\subseteq \bigcup_{p>1} L^p [0,1].$$ Whether the answer is yes or no, I would like to know where to find a proof.

Laura SPD
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    This example is not valid since, for example, $f(x)\in L^{1,1}[0,1]$, so, this function satisfies the statement. – Laura SPD Nov 03 '22 at 01:32
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    Oliver gives a function that is in $L^1$, and not in $L^p$ for $p>2$. Your case is more borderline, the correct example would be something like $1/(x (1+ \log x)^2)$. See https://math.stackexchange.com/questions/55170/is-it-possible-for-a-function-to-be-in-lp-for-only-one-p?noredirect=1&lq=1 – Calvin Khor Nov 03 '22 at 01:38
  • Your idea is good but I’m only considering finite domains while in the other question they are focus on infinite domains. Indeed, the function you propose is not even in $L^1 [0,1]$. – Laura SPD Nov 03 '22 at 01:52
  • @LauraSPD, when responding to someone you should type @ then the person's username. This gives them a notification so that they can respond. I only happened to check this question again to see if you agreed. With regards to the math, I put the square in the wrong place, sorry. It should have been $1/(x(1+(\log x)^2))$. I know that question is on infinite domains, but since the singularity is at a single point, the exact same example works. Said another way, this counterexample on $[0,1]$ trivially extends to a counterexample on $\mathbb R$ – Calvin Khor Nov 03 '22 at 02:25

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