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

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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 '14 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.

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Even worse. You can define a $L^p$-distance but the space is not locally convex. See "Examples of spaces lacking local convexity" in

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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.

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

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I have reverted to the original \Bigl and \Bigr instad of \left and \right. They loor better. – Julián Aguirre Feb 1 '14 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.

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