Editing Lp space (section)

===Embeddings===

Colloquially, if <math>1 \leq p < q \leq \infty,</math> then <math>L^p(S, \mu)</math> contains functions that are more locally singular, while elements of <math>L^q(S, \mu)</math> can be more spread out. Consider the [[Lebesgue measure]] on the half line <math>(0, \infty).</math> A continuous function in <math>L^1</math> might blow up near <math>0</math> but must decay sufficiently fast toward infinity. On the other hand, continuous functions in <math>L^\infty</math> need not decay at all but no blow-up is allowed. More formally:<ref name="VillaniEmbeddings">{{Citation|title=Another note on the inclusion {{math|''L<sup>p</sup>''(''μ'') ⊂ ''L<sup>q</sup>''(''μ'')}}|last=Villani|first=Alfonso|year=1985|journal=Amer. Math. Monthly|volume=92|number=7|pages=485–487|doi=10.2307/2322503|mr=801221|jstor=2322503}}</ref>

#If <math>0<p<q<\infty</math>: <math>L^q(S, \mu) \subseteq L^p(S, \mu)</math> if and only if <math>S</math> does not contain sets of finite but arbitrarily large measure (e.g. any [[finite measure]]). 
#If <math>0<p<q\le\infty</math>: <math>L^p(S, \mu) \subseteq L^q(S, \mu)</math> if and only if <math>S</math> does not contain sets of non-zero but arbitrarily small measure (e.g. the [[counting measure]]).

Neither condition holds for the Lebesgue measure on the real line while both conditions holds for the [[counting measure]] on any finite set. As a consequence of the [[closed graph theorem]], the embedding is continuous, i.e., the [[identity operator]] is a bounded linear map from <math>L^q</math> to <math>L^p</math> in the first case and <math>L^p</math> to <math>L^q</math> in the second. Indeed, if the domain <math>S</math> has finite measure, one can make the following explicit calculation using [[Hölder's inequality]]
<math display="block">\ \|\mathbf{1}f^p\|_1 \leq \|\mathbf{1}\|_{q/(q-p)} \|f^p\|_{q/p}</math>
leading to
<math display="block">\ \|f\|_p \leq \mu(S)^{1/p - 1/q} \|f\|_q .</math>

The constant appearing in the above inequality is optimal, in the sense that the [[operator norm]] of the identity <math>I : L^q(S, \mu) \to L^p(S, \mu)</math> is precisely
<math display="block">\|I\|_{q,p} = \mu(S)^{1/p - 1/q}</math>
the case of equality being achieved exactly when <math>f = 1</math> <math>\mu</math>-almost-everywhere.