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Give a set $E$ and a function $f$ on $E$ such that $\{p:f \in L^p(E)\} =[2,3)$.

This was a homework problem (whose due date has passed) that my brightest fellow classmates AND the class TA weren't able to solve. The hint given to us was to analyze $$f(x)=\frac{1}{x^\alpha \log^{\beta}(x)}$$ for various $\alpha$ and $\beta$, on $(0,\frac{1}{2}]$ and on $[2, \infty)$.

I tried looking at the function $f(x)$ given in the hint and noted that $\int_2^\infty f<\infty$ if and only if $\alpha > 1$ since if $\beta>0$ we can bound the integral by $\log^\beta(2) \int_2^\infty f <\infty$ but if $\alpha <1$ then $x^\epsilon\geq\log^\beta(x)$ for all $x\geq c \in \mathbb{R}$ where $\epsilon$ is such that $\alpha + \epsilon < 1$, hence $$\int_2^\infty f \geq \int_c^\infty x^{(-\alpha - \epsilon)}\geq \int_c^\infty \frac{1}{x} = \infty.$$ I also noted $$\int_2^\infty \frac{1}{x\log(x)} = \lim_{x \to \infty}\log(\log(x))-\log(\log(2))=\infty$$ and that $\int_2^\infty \frac{1}{x\log^\beta(x)}$ diverges for $\beta<1$ and converges for $\beta>1$.

A similar analysis on $(0, \frac{1}{2}]$ shows convergence if $\alpha<1$ and, if $\alpha=1,$ for $\beta<1$, and divergence elsewhere.

I was not able to reach the conclusion from this information, and am pretty stumped after hours of searching and testing various combinations of the above function integrated over the indicated sets.

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1 Answer 1

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Let $f_{\alpha,\beta}(x) = \frac{1}{x^\alpha (\log x)^\beta}$. You have shown that $f_{\alpha,\beta} \in L^p([2, \infty))$ whenever:

  • $p > 1/\alpha$, or
  • $p = 1/\alpha$ and $\beta > \alpha$.

So, in particular, $\{p : f_{1/2,1} \in L^p([2,\infty))\}=[2, \infty]$.

You have also shown that $f_{\alpha,\beta} \in L^p([0,1/2])$ whenever:

  • $p < 1/\alpha$, or
  • $p = 1/\alpha$ and $\beta < \alpha$.

So, in particular, $\{p : f_{1/3,0} \in L^p([0, 1/2])\}=[1, 3)$.

Let $$ g(x)=\begin{cases} f_{1/2,1}(x) & x > 2 \\ f_{1/3,0}(x-1) & \frac{3}{2} < x < 2 \end{cases} $$ Then $\|g\|_{L^p([3/2,\infty))}=\|f_{1/2,1}\|_{L^p([2,\infty))}+\|f_{1/3,0}\|_{L^p([0,1/2])}$ for all $p$; thus $g \in L^p([3/2,\infty))$ whenever $p \in [1,3) \cap [2, \infty] = [2, 3)$.

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