TITLE: $f \in L^1$, but $f \not\in L^p$ for all $p > 1$
QUESTION [12 upvotes]: "Find an $f \in [0,1]$ such that $f \in L^1$ but $f \not\in L^p$ for any $p > 1$."
I've thought about doing something like $$f(x) = \frac{1}{x}$$ where $|f|^p = \frac{1}{x^p}$ doesn't converge when $p > 1$.  But this function isn't itself in $L^1$.  Could someone please give me a hint for how to solve this problem?  I wish there were a situation where you had convergence on the closed half disc $[0,1]$ and divergence on $(1, \infty)$, rather than my current predicament where I have convergence on the open half-disc $[0, 1)$ and divergence on $[1, \infty)$.

REPLY [17 votes]: For $x\in (0,1), $ define
$$f(x) = \frac{1}{x(|\ln x|+1)^2}.$$
Then $f \in L^1(0,1),$ but $f \notin L^p(0,1)$ for all $p>1.$