# Are $L^\infty(\Bbb R)$ and $L^2(\Bbb R)$ homeomorphic?

It's easy to see that, for $1\le p,q< \infty$ the spaces $L^p(\Bbb R)$ and $L^q(\Bbb R)$ of $p$-th and $q$-th power integrable functions on the real line are homeomorphic as topological spaces. In fact, the map $f(x)\mapsto sgn(f(x))|f(x)|^{q/p}$ provides an explicit homeomorphism $L^p(\Bbb R) \to L^q(\Bbb R)$.

However this argument cannot be applied to the case where $p$ or $q$ equals infinity. Thus, I'm asking whether there is a homeomorphism from $L^\infty(\Bbb R)$ to any (and hence every) $L^p(\Bbb R)$ with $p<\infty$.

• Isn't $L^2(\mathbb{R})$ separable and $L^\infty(\mathbb{R})$ nonseparable? – Paul McKenney Feb 9 '15 at 22:38
• A more interesting question would be: are $L^2({\bf R})$ and $C_0({\bf R})$ homeomorphic? – Yemon Choi Feb 9 '15 at 23:07
• Any two separable infinite dimensional Frechet spaces are homeomorphic. This is due to M. I. Kadec. – Bill Johnson Feb 9 '15 at 23:25
• The $q/p$ should be $p/q$ if you want it to take $L^p$ to $L^q$. – Robert Israel Feb 10 '15 at 2:24
• @PaulMcKenney I like the use of the rhetorical question... – Yemon Choi Feb 10 '15 at 12:18

Paul is right. $L^2(\mathbb{R})$ is separable. (The rational simple functions ought to be one example of something ctbl. and dense.)
However, $L^{\infty}(\mathbb{R})$ isn't separable. By Jones' Lemma, if $L^{\infty}(\mathbb{R})$ were separable then any closed discrete (i.e., nonclustering) set must be of size less than continuum. But the characteristic functions $\chi_{[0,x]}$, $x\in[0,\infty)$, are pairwise of distance 1 from eachother.
• Jones' Lemma seems like overkill here. Since the continuum-many functions $\chi_{[0,x]}$ are pairwise distance 1 in $L^\infty$, the continuum-many open balls $B(\chi_{[0,x]}, 1/2)$ are pairwise disjoint. Any dense set must contain points in every one of these balls, so a dense set in $L^\infty$ must have size at least continuum. – Nate Eldredge Jan 4 '17 at 20:14