$L^1$ function that is not $L^p$ for $p > 1$.

Question

I ran into a rather interesting problem recently, which asks for a function $f \in L^1(\mathbb{R})$ such that $f \not\in L^p(\mathbb{R})$ for all $p > 1$. My initial thought was that $f(x) = x^{-\alpha}$ is integrable on $[0,1]$ if and only if $\alpha < 1$. So I figured I could get a solution by pasting together functions of the form $f_n(x) = (x - c_n)^{-\alpha_n}$, where $\alpha_n \to 1, \alpha_n < 1$. I came up with: $$f(x) = \sum_{n=0}^{\infty}\frac{1}{2^n}\chi_{(n,n+1]}(x - n)^{-\tfrac{n}{n + 1}}.$$ The factor of $2^{-n}$ and the fact that $\alpha_n < 1$ for all $n$ ensures that $f \in L^1(\mathbb{R}$). However, if $p > 1$, then there exists some $N$ such that $$p\frac{N}{N + 1} \geq 1,$$ thus $f \not\in L^p(\mathbb{R}^d)$. Is this correct? I would also be interested to see other such functions.

Answer 1

Your solution looks correct to me. Another (somewhat simpler) example is the function $$ f(x)=x^{-1}\log(x)^{-2}\chi_{(0,\frac{1}{2}]}(x) $$

Answer 2

I think your example is correct, but way too complicated. There are two things that need to be balanced for integrability on $\mathbb{R}$, that is integrability of singularities and sufficiently fast decay at infinity.

Taking higher powers of your function (as is the case with $L^p$-norms) improves your decay at infinity. So the crux of the matter is integrability of local singularities. carmichael561's suggestion here is spot on.