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I have seen a theorem that assures that the set $A=\lbrace p\in [1,\infty]: u\in L^p(0,\infty)\rbrace$ is an interval. It is easy to find a function $u$ for which $A$ is the empty-set or $[1,\infty]$ or $\lbrace \infty \rbrace$. Also, the function $$\frac{1}{x^{1/a}[\log^2(x)+1]}$$ is $L^p(0,\infty)$ iff $a=p$ (Is it possible for a function to be in $L^p$ for only one $p$?). I could find a function $u$ s.t. $A$ is of the form $(a,\infty ]$ and $[1,a)$.

I wonder if there exists a function $u$ for which the interval $A$ is of the form $[a,b]$ or $(a,b)$ or $(a,b]$ or $[a,b)$ for $1 \leq a<b \leq \infty$. This might be hard, I do not expect a complete answer but any idea or hint would be appreciated.

Thanks in advance.

Bernard
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Senna
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1 Answers1

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The function $f(x) = x^{-1/b} \chi_{(0,1)}(x)$ belongs to $L^p(0,\infty)$ if and only if $p < b$.

The function $g(x) = x^{-1/a} \chi_{(1,\infty)}(x)$ belongs to $L^p(0,\infty)$ if and only if $p > a$.

The function $f+g$ belongs to $L^p(0,\infty)$ if and only if $p \in (a,b)$.

The case of closed or half-closed intervals is a little harder but similar.

Umberto P.
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  • Thanks! Do you know where can I find the closed cases? – Senna Mar 19 '19 at 19:50
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    I had hoped you'd give it a try yourself. Here is a hint: if $a_k \nearrow a$ and $g_k = x^{-1/a_k} \chi_{(1,\infty)}(x)$ for each $k$, try to use a sum of the form $g = \sum c_k g_k$ where the $c_k$ are constant. For $g$ to belong to $L^p$ it is necessary that $p > a_k$ for all $k$. If you choose the $c_k$ properly $p \ge a$ will be sufficient too. – Umberto P. Mar 19 '19 at 20:18
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    For the closed intervals, one can use also logs, like $h(x) = x^c , \log(x)^d$ for suitable exponents $c$ and $d$. – gerw Mar 20 '19 at 06:35