Lp space
L (0 < p < 1)

Let (S, Σ, μ) {\displaystyle (S,\Sigma,\mu)} be a measure space. If 0 < p < 1, {\displaystyle 0<p<1,} then L p (μ) {\displaystyle L^{p}(\mu)} can be defined as above: it is the quotient vector space of those measurable functions f {\displaystyle f} such that N p (f) = ∫ S | f | p d μ < ∞. {\displaystyle N_{p}(f)=\int _{S}|f|^{p}\,d\mu <\infty.}

As before, we may introduce the p {\displaystyle p} -norm ‖ f ‖ p = N p (f) 1 / p, {\displaystyle \|f\|_{p}=N_{p}(f)^{1/p},} but ‖ ⋅ ‖ p {\displaystyle \|\cdot \|_{p}} does not satisfy the triangle inequality in this case, and defines only a quasi-norm. The inequality (a + b) p ≤ a p + b p, {\displaystyle (a+b)^{p}\leq a^{p}+b^{p},} valid for a, b ≥ 0, {\displaystyle a,b\geq 0,} implies that (Rudin 1991, §1.47) N p (f + g) ≤ N p (f) + N p (g) {\displaystyle N_{p}(f+g)\leq N_{p}(f)+N_{p}(g)} and so the function d p (f, g) = N p (f − g) = ‖ f − g ‖ p p {\displaystyle d_{p}(f,g)=N_{p}(f-g)=\|f-g\|_{p}^{p}} is a metric on L p (μ). {\displaystyle L^{p}(\mu).} The resulting metric space is complete; the verification is similar to the familiar case when p ≥ 1. {\displaystyle p\geq 1.} The balls B r = { f ∈ L p: N p (f) < r } {\displaystyle B_{r}=\{f\in L^{p}:N_{p}(f)<r\}} form a local base at the origin for this topology, as r > 0 {\displaystyle r>0} ranges over the positive reals. These balls satisfy B r = r 1 / p B 1 {\displaystyle B_{r}=r^{1/p}B_{1}} for all real r > 0, {\displaystyle r>0,} which in particular shows that B 1 {\displaystyle B_{1}} is a bounded neighborhood of the origin; in other words, this space is locally bounded, just like every normed space, despite ‖ ⋅ ‖ p {\displaystyle \|\cdot \|_{p}} not being a norm.

In this setting L p {\displaystyle L^{p}} satisfies a reverse Minkowski inequality, that is for u, v ∈ L p {\displaystyle u,v\in L^{p}} ‖ | u | + | v | ‖ p ≥ ‖ u ‖ p + ‖ v ‖ p {\displaystyle {\Big \|}|u|+|v|{\Big \|}_{p}\geq \|u\|_{p}+\|v\|_{p}}

This result may be used to prove Clarkson's inequalities, which are in turn used to establish the uniform convexity of the spaces L p {\displaystyle L^{p}} for 1 < p < ∞ {\displaystyle 1<p<\infty } (Adams & Fournier 2003).

The space L p {\displaystyle L^{p}} for 0 < p < 1 {\displaystyle 0<p<1} is an F-space: it admits a complete translation-invariant metric with respect to which the vector space operations are continuous. It is the prototypical example of an F-space that, for most reasonable measure spaces, is not locally convex: in ℓ p {\displaystyle \ell ^{p}} or L p (), {\displaystyle L^{p}(),} every open convex set containing the 0 {\displaystyle 0} function is unbounded for the p {\displaystyle p} -quasi-norm; therefore, the 0 {\displaystyle 0} vector does not possess a fundamental system of convex neighborhoods. Specifically, this is true if the measure space S {\displaystyle S} contains an infinite family of disjoint measurable sets of finite positive measure.

The only nonempty convex open set in L p () {\displaystyle L^{p}()} is the entire space (Rudin 1991, §1.47). As a particular consequence, there are no nonzero continuous linear functionals on L p (); {\displaystyle L^{p}();} the continuous dual space is the zero space. In the case of the counting measure on the natural numbers (producing the sequence space L p (μ) = ℓ p {\displaystyle L^{p}(\mu)=\ell ^{p}}), the bounded linear functionals on ℓ p {\displaystyle \ell ^{p}} are exactly those that are bounded on ℓ 1, {\displaystyle \ell ^{1},} namely those given by sequences in ℓ ∞. {\displaystyle \ell ^{\infty }.} Although ℓ p {\displaystyle \ell ^{p}} does contain non-trivial convex open sets, it fails to have enough of them to give a base for the topology.

The situation of having no linear functionals is highly undesirable for the purposes of doing analysis. In the case of the Lebesgue measure on R n, {\displaystyle \mathbb {R} ^{n},} rather than work with L p {\displaystyle L^{p}} for 0 < p < 1, {\displaystyle 0<p<1,} it is common to work with the Hardy space H whenever possible, as this has quite a few linear functionals: enough to distinguish points from one another. However, the Hahn–Banach theorem still fails in H for p < 1 {\displaystyle p<1} (Duren 1970, §7.5).
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