7

This question was on a problem set regarding $L^p$ spaces in an undergraduate-level real analysis course. I actually used an answer on StackExchange to help me provide an example, but I couldn't provide adequate justification for why it works. The answer I used is linked here. I modified this function a little to make it so that it only converges if $p = 2$, namely I defined my function as follows: $$f(x)=\frac{1}{x^{2/p} \ln^2(x^{2/p})}$$

But I can't really explain why it works (and to be honest, I'm not really sure it does work). My professor hinted in class that there is a solution that involves a continuous piecewise function agreeing at $x=1$ that is basically a generalization of the observation that for the following functions: $$g(x)=\begin{cases} \frac{1}{\sqrt{x}} & 0<x<1 \\ 0 & \text{otherwise}\end{cases}$$ $$h(x)=\begin{cases} \frac{1}{x} & x>1 \\ 0 & \text{otherwise}\end{cases}$$ we can easily show that $g \in L^1(\mathbb{R}), \, g \not\in L^2(\mathbb{R})$ and $h \in L^2(\mathbb{R}), \, h \not\in L^1(\mathbb{R})$. So I would imagine that the solution based on this hint would involve some sort of clever manipulation of powers to incorporate all $p \in [1, 2) \cup (2, \infty]$ somehow, but I have no idea how to do that. I'm pretty lost with this problem. What is the best way to go about it?

Community
  • 1
ProXy
  • 301
  • 2
  • 8

1 Answers1

1

No, $f(x)=\frac{1}{x^{2/p} \ln^2(x^{2/p})}$ does not work by itself. The answer you linked does not contain an explicit example of such a function: it gives one piece of it, and then says "stitch this function together with its inverse".

You'd be better off using Robert Israel's answer instead: $$f(x) = \frac{1}{x^{1/p} (\ln(x)^2+1)} \qquad \text{on} \qquad (0, \infty)$$ You don't need to modify this for $L^2$ other than setting $p=2$.

  • Convergence of the integral of $f^p$ near $0$ and $\infty$ follows by comparison $f(x)^p \le \frac{1}{x \ln(x)^{2p}}$ which can be integrated directly.
  • Divergence of the integral of $f^q$ with $q\ne p$ follows from the fact that the factor of $x^{q/p}$ in the denominator has exponent different from $1$, so it lands on the "divergent" side of $1$ either at $0$ or at $\infty$. The logarithmic term can be estimated by $\ln(x)^2+1 \le Cx^\epsilon$ at $\infty$, or $\ln(x)^2+1 \le Cx^{-\epsilon}$ near $0$.
Community
  • 1