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Let $X$ be the interval $[0,1]$ with Lebesgue measure. Is there a function $f\in L^p(X)$ for all $p\in[1,\infty)$ that is not $\in L^\infty(X)$? If so, what is an example?

Motivation: In a course on measure theory this fall, I've learned proofs that $L^p(X)\supset L^q(X)$ if $q>p$ and that if $f\in L^\infty(X)$, then $\|f\|_\infty = \lim \limits_{p\to\infty} \|f\|_p$. This prompted me to wonder if $L^\infty(X) = \bigcap _p L^p(X)$. A classmate gave me a general theoretical reason to believe the contrary: $L^p(X)$ is a reflexive space for $1 \lt p \lt \infty$ but not for $p=1,\infty$; but intersections of reflexive spaces are reflexive. This logic seems sound to me; but it implies the containment $L^\infty(X) \subset \bigcap_p L^p(X)$ is strict. If so, there must be a function that is $L^p$ for all $p$ but not a.e. bounded. What is it?

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    $\begingroup$ There is a flaw in your classmate's theoretical argument: If you have closed reflexive subspaces of a Banach space then their intersection is reflexive (simply because closed subspaces of reflexive spaces are reflexive). However, $L^q(X)$ is dense in $L^p(X)$ for $ \infty \geq q \gt p \geq 1$, so you cannot conclude. $\endgroup$ – t.b. Dec 12 '11 at 3:27
  • $\begingroup$ To complement Jonas's answer below: if you merely want a concrete example on some probability space $(\Omega,\mu)$, then a Gaussian random variable, viewed as a function $\Omega\to{\mathbb R}$, is in every $L^p(\Omega,\mu)$ -- it ``has moments of all orders'' -- but is not a.s. bounded. $\endgroup$ – user16299 Apr 4 '12 at 6:40
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The logarithm is an example.

To see that $\log$ is in $L^p$ for each $p>0$, it suffices to see that $|\log(t)|^p<t^{-1/2}$ for $t$ sufficiently small. This follows from the fact that $\lim\limits_{t\searrow0}\;|\log(t)|t^{1/(2p)}=0$, which is a quick application of l'Hôpital's rule.

The $L^p$ norm of $\log$ happens to be $\Gamma(p+1)^{1/p}$. This can be seen by making the change of variables $t=-\log(u)$ in the integral $\Gamma(x)=\int\limits_{0}^\infty t^{x-1}e^{-t}\:dt$. So by Stirling's approximation, $\|\log\|_p$ is close to $p/e$ when $p$ is large.

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For another example, observe that the series $$\sum_{n=1}^\infty n^p\frac{1}{2^n}$$ converges for any $p$ from root test since $\limsup_{n\rightarrow \infty} \left(\frac{n^p}{2^n}\right)^{1/n} = \frac{1}{2} < 1$.

You can define the function $$f(x) = \sum_{n=1}^\infty n\chi_{(2^{-n}, 2^{-n+1}]}(x)$$ as an example.

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    $\begingroup$ +1 but note that morally this is the same example as Jonas' - observe that your $f$ is bounded by constant multiples of $-\log x$. $\endgroup$ – Ben Blum-Smith Jan 23 '16 at 22:52
  • $\begingroup$ But how do we know that $\displaystyle \left(\sum_{n=1}^\infty \frac{n}{2^n} \right)^p= \sum_{n=1}^\infty \frac{n^p}{2^n}$ $\endgroup$ – Twink Apr 11 '20 at 23:03
  • $\begingroup$ @Twink the two terms are not equal, if you compute the $L^p$ norm of $f$, you get $\sum_{n=1}^\infty \frac{n^p}{2^n}$. $\endgroup$ – Xiao Apr 12 '20 at 16:06
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Xiao and Jonas Meyer give good answers, but there maybe somethings wrong with your motivation:

$L^p(X)\supset L^q(X)$ if $q<p$.

Because I saw a proposition in Folland's Real Analysis which on P186

Suppose $0<p<q<\infty$. Then $L^p \nsubseteq L^q $ iff $X$ contains sets of arbitrarily small positivite measure, and $L^q \nsubseteq L^p $ iff $X$ contains sets of arbitrarily large finite measure.

So, in general we have $L^p \nsubseteq L^q$ for all $p\neq q$.

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  • $\begingroup$ This should be left as a comment to the original post, not an answer. $\endgroup$ – Alex Dec 19 '19 at 11:27

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