# A function “almost not” in $L^1$

Let $f$ be a measurable function on a finite measure space. If $f \notin L^p$ for all $p > 1$, is it true that $f \notin L^1$?

Unfortunately the inequality I am using to prove that $f \notin L^p$ does not work for $p = 1$, but I'm hoping for some kind of "continuity of the norm" argument whereby $\lVert f \rVert_1 = \lim_{p \to 1} \lVert f \rVert_p$ would imply that $\lVert f \rVert_1 = \infty$, but obviously the limit argument is not defined.

• It is not true. There are $L^1$ functions that are not integrable in any higher $L^p$, just as there are $L^p$ functions that are not in any $L^{p + \epsilon}$. – T. Bongers Jul 31 '16 at 23:37
• at.yorku.ca/cgi-bin/… – T. Bongers Jul 31 '16 at 23:39

## 2 Answers

An explicit exemple of a function in $L^1 ([0,\frac{1}{e}])$ for that it's not in $L^p$ for $p>1$ :

$$f(x) = \frac{1}{x \ln(x)^2}$$

• I love explicit counterexamples -- thanks! – user369210 Aug 4 '16 at 19:01

Let $p_n = 1 + 1/n$, and let $f_n \in L^1 \setminus L^{p_n}$, $f_n \geq 0$. Let $$f = \sum\limits_{n=1}^\infty {1 \over 2^n \|f_n\|_{L^1}} f_n.$$ Then $f \in L^1$ by the monotone convergence theorem, but $f \notin L^{p_n}$ for any $n$.