Take the 2-minute tour ×
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.

Why do we consider Lebesgue spaces for $p$ greater than and equal to $1$ only and not for $p$ any real number?

share|improve this question
1  
James Roberts considered such $L_p$-spaces, $0<p<1$ as was able to construct in each, a compact convex set without extreme points; thus resolving a long standing problem in Functional Analysis. The techniques used in this construction, arguably, led to the resolvement of other problems concerning $L_1$. –  David Mitra Jan 23 at 16:19

5 Answers 5

The reason is that we would like to define a norm by the following formula: $$ \|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}. $$ Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for $p\geq 1$. See more details in Chapter $6$ of Folland's book: Real Analysis.

share|improve this answer

Even worse. You can define a $L^p$-distance but the space is not locally convex. See "Examples of spaces lacking local convexity" in http://en.wikipedia.org/wiki/Locally_convex_topological_vector_space.

share|improve this answer

As a concrete example, if we consider the domain $[0,2]$ and try to set $p=1/2$, then our "norm" would be $$ \|f\|_{1/2} = \left( \int_0^2 \sqrt{|f(x)|} \, dx\right)^2 $$

Consider then

$$ f(x) = \begin{cases} 1 & x<1 \\ 0 & x\ge 1 \end{cases} \qquad g(x) = \begin{cases} 0 & x<1 \\ 1 & x\ge 1 \end{cases}$$

We then have $\|f\|_{1/2} = \|g\|_{1/2} = 1$ but $\|f+g\|_{1/2} = 4$, violating the triangle inequality.

share|improve this answer

Mainly because $$ \Bigl(\int_D|f|^p\Bigr)^{1/p} $$ is not a norm if $0<p<1$.

share|improve this answer
    
I have reverted to the original \Bigl and \Bigr instad of \left and \right. They loor better. –  Julián Aguirre Feb 1 at 14:50

As it is already mentioned, $$ \|f\|_p=\left(\int_X |f|^p\,dx\right)^{1/p}, $$ is a norm iff $p\ge 1$, since only for $p\ge 1$ satisfies the triangle inequality.

Nevertheless, the spaces $L^p(X)$, for $p\in (0,1)$ are quite interesting since they serve as the most typical example for non locally convex topological vector spaces.

share|improve this answer

Your Answer

 
discard

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.