{"id":97,"date":"2008-08-28T20:37:33","date_gmt":"2008-08-28T19:37:33","guid":{"rendered":"https:\/\/wpethzprd.ethz.ch\/kowalski\/2008\/08\/28\/can-lp-be-the-same-as-lq\/"},"modified":"2024-02-20T13:40:58","modified_gmt":"2024-02-20T11:40:58","slug":"can-lp-be-the-same-as-lq","status":"publish","type":"post","link":"https:\/\/blogs.ethz.ch\/kowalski\/2008\/08\/28\/can-lp-be-the-same-as-lq\/","title":{"rendered":"Can L^p be the same as L^q?"},"content":{"rendered":"

A fairly standard question that students are asked after a course in real analysis (involving Lebesgue integration) is to show that there is no inclusion between Lp<\/sup><\/em> spaces on R<\/strong><\/em> with respect to Lebesgue measure. This is usually done (sometimes by hinting or prompting) by considering for which values of \u256c\u2592<\/em> and p<\/em> the functions<\/p>\n

$latex x\\mapsto x^{\\alpha}$<\/p>\n

are in Lp<\/sup><\/em> (after truncating either for |x|<1<\/em> or for |x|>1<\/em>, to avoid divergences).<\/p>\n

Once I was present as observer at an oral examination for such a course, and the professor in charge had raised this question more or less by asking<\/p>\n

Is it possible that Lp<\/sup><\/em> is the same as Lq<\/sup><\/em>, if p<\/em> is not equal to q<\/em>?<\/p><\/blockquote>\n

where the meaning was clearly, implicitly, the one above; and I wondered (aloud) if the answer was the same in a more abstract way: is it possible that Lp<\/sup>(X,\u256c\u255d)<\/em> be isomorphic<\/em> to Lq<\/sup>(X,\u256c\u255d)<\/em> if p<\/em> is different from q<\/em>? This is a question that makes sense for general measure spaces, of course, and one where one must be careful to specify what “isomorphic” is taken to mean. In the algebraic sense, this is a question of cardinality only, and doesn’t seem very interesting, but isomorphism as topological vector spaces seems much more natural. Note that the answer is not immediately clear for the classical Lebesgue spaces over R<\/strong><\/em>: even if there is no inclusion, there might exist some clever linear renormalization map that sends (for instance) a square-integrable function bijectively — and continuously — to an integrable one.<\/p>\n

But still, the answer is (not surprisingly) “No”, provided the only obvious reservation is made: if Lp<\/sup><\/em> and Lq<\/sup><\/em> are finite dimensional, then they are isomorphic as topological vector spaces, as is well known (they are not necessarily isometric, of course). But otherwise, we have<\/p>\n

Theorem.<\/strong> Let (X,\u256c\u255d)<\/em> be a measure space, and let p<\/em> and q<\/em> be real numbers at least 1 such that Lp<\/sup>(X,\u256c\u255d)<\/em> has infinite dimension. Then Lp<\/sup>(X,\u256c\u255d)<\/em> and Lq<\/sup>(X,\u256c\u255d)<\/em> are isomorphic, as topological vector spaces, if and only if p=q<\/em>.<\/p><\/blockquote>\n

I think this goes back to Banach, at least for the classical spaces of functions on R<\/strong><\/em> (if I understand correctly Chapter XII of his book Th\u00e9orie des op\u00e9rations lin\u00e9aires<\/a>). For the general case, although I didn’t find a reference, this must be well-known since it is a direct consequence of the computation of the type<\/em> and cotype<\/em> invariants of such Banach spaces. (The only reason I actually had the idea to look at these is that I was browsing pleasurably into the nice book of Li and Queff\u00e9lec<\/a> on the geometry of Banach spaces, where type and cotype are described in detail; this is overall very far from what I can claim any expertise to…)<\/p>\n

Indeed, for an infinite dimensional Banach space of the form Lp<\/sup>(X,\u256c\u255d)<\/em>, it is known that<\/p>\n

$latex type(L^p(X,\\mu))=\\min(2,p),\\ \\ \\ \\ \\ \\ cotype(L^p(X,\\mu))=\\max(2,p)$<\/p>\n

where the (best) type<\/em> (denoted type(E)<\/em>) of a Banach space E<\/em> is defined to be the largest real number p<\/em> such that<\/p>\n

$latex \\int_0^1{||\\sum_{j=1}^n{r_j(t)x(j)}||dt}\\leq M\\left(\\sum_{j=1}^n{||x(j)||^p}\\right)^{1\/p}$<\/p>\n

for any n>1<\/em>, any finite sequence of vectors x(j)<\/em> in E<\/em>, and some constant M>0<\/em> (independent of n<\/em>), where<\/p>\n

$latex r_j(t)=\\mathrm{sign}\\sin 2^j\\pi t$<\/p>\n

denotes the sequence of Rademacher functions<\/em><\/a>. Dually, the cotype is the smallest real number q<\/em> such that<\/p>\n

$latex \\int_0^1{||\\sum_{j=1}^n{r_j(t)x(j)}||dt}\\geq m\\left(\\sum_{j=1}^n{||x(j)||^q}\\right)^{1\/q}$<\/p>\n

for some constant m>0<\/em>. The results above on the type and cotype of Lp<\/sup><\/em> spaces are explained in Section III.3 of the book of Li and Queff\u00e9lec.<\/p>\n

These definitions show that the type and cotype are preserved under continuous linear isomorphisms, so if we have infinite-dimensional spaces Lp<\/sup>(X,\u256c\u255d)<\/em> and Lq<\/sup>(X,\u256c\u255d)<\/em> which are isomorphic, their types and cotypes must coincide, i.e., we must have<\/p>\n

$latex \\min(2,p)=\\min(2,q),\\ \\ \\ and\\ \\ \\ \\max(2,p)=\\max(2,q),$<\/p>\n

which means p=q<\/em>.<\/p>\n","protected":false},"excerpt":{"rendered":"

A fairly standard question that students are asked after a course in real analysis (involving Lebesgue integration) is to show that there is no inclusion between Lp spaces on R with respect to Lebesgue measure. This is usually done (sometimes by hinting or prompting) by considering for which values of \u256c\u2592 and p the functions … Continue reading Can L^p be the same as L^q?<\/span><\/a><\/p>\n","protected":false},"author":625,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-97","post","type-post","status-publish","format-standard","hentry","category-blogroll"],"_links":{"self":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts\/97","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/users\/625"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/comments?post=97"}],"version-history":[{"count":0,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/posts\/97\/revisions"}],"wp:attachment":[{"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/media?parent=97"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/categories?post=97"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ethz.ch\/kowalski\/wp-json\/wp\/v2\/tags?post=97"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}