Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Lieb and Loss's Analysis, I saw that they mentioned $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $ L^2(\mathbb{R}^n)$ (dense wrt the $L^2$ norm, I think). But I didn't find its proof in the book. So I wonder why it is?

Does this conclusion hold for $L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)$ for any measure space $(\Omega, \mathcal{F}, \mu)$?

Are there similar statements if replace $L^1$ with $L^p$, and $ L^2$ with $L^q$ for $p \leq q \in (0, \infty ]$ or $\in [1, \infty]$?

Thanks and regards!

share|cite|improve this question
Hint: Consider simple functions. – Martin Dec 27 '12 at 18:21
The intersection contains all continuous functions with compact support, or all integrable simple functions, or... – Mariano Suárez-Alvarez Dec 27 '12 at 18:21
The set of smooth functions with compact support is dense in either. – copper.hat Dec 27 '12 at 18:30
@Tim, that might be a good signal that you need to review Lebesgue spaces! – Mariano Suárez-Alvarez Dec 27 '12 at 18:35
The general version is also true, unless $q=\infty$. As an exercise, you may want to prove that $L^p(\mathbb R)\cap L^\infty (\mathbb R)$ is not dense in $L^\infty (\mathbb R)$ for any finite $p$. (Hint: consider the function $f$ that is identically $1$ and show that $\|f-g\|_{L^\infty}\le 1/2$ implies that $g$ is not integrable). – user53153 Dec 27 '12 at 19:12
up vote 11 down vote accepted

Let me collect all the comments into a single answer.

Since the space of integrable simple functions is dense in $L^p(\Omega, \mathcal{F}, \mu)$ for all $p\in[1,+\infty)$, and in addition, all simple functions are in $L^\infty(\Omega, \mathcal{F}, \mu)$, we have that $\bigcap_{1\leq p\leq\infty}L^p(\Omega, \mathcal{F}, \mu)$ is dense in $L^q(\Omega, \mathcal{F}, \mu)$ for all $q\in[1,+\infty)$.

If $q=\infty$ the answer depends on whether or not $\mu$ is finite.

  • If $\mu$ is finite, then $L^\infty(\Omega, \mathcal{F}, \mu)\subseteq L^p(\Omega, \mathcal{F}, \mu)$ for each $p$, so obviously we can put $q=\infty$ above.
  • If $\mu$ is infinite (like the Lebesgue measure), then any nonzero constant function is in $L^\infty$, but very far from any integrable function (since no integrable function can be bounded away from zero on a set of infinite measure).

Worth mentioning, if inessential for this exercise, is that if $\Omega,\mu$ are sufficiently well-behaved ($\Omega$ is locally compact Hausdorff, $\mu$ is inner and outer regular, locally finite), then compactly supported continuous functions are dense in each $L^p(\mu)$ with $p<\infty$. This applies in particular to Haar measures on locally compact Hausdorff groups, such as the Lebesgue measure.

share|cite|improve this answer
Thanks! @tomasz: I think $L^\infty(\Omega, \mathcal{F}, \mu)\supseteq L^p(\Omega, \mathcal{F}, \mu), \forall p \in (0, \infty)$? My reason is $L^\infty$ is defined as the set of measurable functions that are bounded up to a set of measure zero, and if $f \notin L^\infty$, then there exists a subset of measure nonzero on which $|f|$ is $\infty$, so $f \notin L^p, \forall p \in (0, \infty)$. So I wonder why "If $\mu$ is finite, then $L^\infty(\Omega, \mathcal{F}, \mu)\subseteq L^p(\Omega, \mathcal{F}, \mu)$ for each $p$"? If we both are right, then If $\mu$ is finite, $L^\infty = L^p$? – Tim Dec 28 '12 at 0:19
I posted my last comment as a new question here – Tim Dec 28 '12 at 0:47
@Tim: No. For example, $\log(x)$ is in $L^p([0,1],\mathcal B,\lambda)$ for all $p<\infty$, but not in $L^\infty$. – tomasz Dec 28 '12 at 2:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.