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Chapter 5
The Lp^ spaces
In this chapter we shall give an introduction to the Lp^ spaces which plays an importan rolle in mathematics. Particularly,tions. L^2 who is an inner product space with many applica-
5.1 Definition and Elementals properties of the Lp^ spaces.
Definition 5.1. Let p ∈ R with 1 ≤ p < ∞, we shall define
LP(Ω) =
f : Ω −→ R : f is measurable and
Z Ω^ |^ f^ |dm^ <^ ∞
We do not distinguish between two functions that are equal c.t.p. Here Ω ⊆ R is a measurable subset of R. We shall denote ∥ f ∥LP =
Z
Ω^ |^ f^ |Pdm
1/P
Soon, we will prove that ∥ f ∥LP is a norm in LP(Ω). Definition 5.2. L∞(Ω) = { f : Ω −→ R : f is measurable and there exists C ∈ R such that | f (x)| ≤ C a.e}, and we denote
∥ f ∥L∞^ = inf{C : | f (x)| ≤ C, a.e in Ω}. We will see later that ∥ · ∥L∞^ is a norm. 104
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Remark 5.3. If f ∈ L∞(Ω), then | f (x)| ≤ ∥ f ∥L∞^ , a.e. In fact: There exists a sequence {Cn}∞ n= 1 of non negative numbers such that
n^ lim→∞ Cn^ =^ ∥^ f^ ∥L∞^ , and | f (x)| ≤ Cn, a.e., which means that there is a set En ⊂ Ω with m(En) = 0 such that | f (x)| ≤ Cn, x ∈ Ω\En. Then, if we put E = n[^ ∞= 1 En, we obtain that | f (x)| ≤ Cn, x ∈ Ω\E and m(E) = 0. So, passing to the limit n goes to infinite, we get that | f (x)| ≤ ∥ f ∥L∞^ , a.e.
Notation: Let 1 ≤ p ≤ ∞. We define P′^ by the following relation 1 p +^
p′^ =^ 1,^ if^ p^ >^ 1. If p = 1, then p′^ = ∞.
Theorem 5.1.1. ( H older Inequality ¨ ) Let f ∈ LP(Ω), g ∈ LP′ (Ω) with 1 ≤ p ≤ ∞. Then f .g ∈ L^1 (Ω) and Z Ω^ |^ f g|dm^ ≤ ∥^ f^ ∥LP^ ∥g∥LP′^.^ (5.1) Proof Recordemos la desigualdad de Young. The conclusion is trivial if p = 1 or p = ∞. Supongamos entonces que 1 < p < ∞.
ab ≤ (^1) P aP^ + (^) P^1 ′ bP′ , a ≥ 0, b ≥ 0 (5.2) En efecto, la funci ´on logaritmo es concava en [0, ∞) (convexa hacia abajo)
log
P aP^ +^
P′^ bP
′^ ≥ 1
P log^ aP^ +^
P′^ log^ bP
′
= log(a) + log b = log(ab)
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■
Ejercicio. (Exercise) Suppose that fi ∈ LPi^ , i = 1,... , k with 1 P =^
P 1 +^
P 2 +^ · · ·^ +^
Pk^ ≤^1
then f = f 1. f 2... fk ∈ LP^ and
∥ f ∥LP ≤ ∥ f 1 ∥LP 1... ∥ fk∥LPk.
it follows thatIn particular, if^ f^ ∈^ LP^ ∩^ Lq^ with 1^ ≤^ P^ ≤^ q^ ≤^ ∞, then^ f^ ∈^ Lr^ ∀^ r^ >^ 0,^ ∋^ p^ ≤^ r^ ≤^ q
∥ f ∥r ≤ ∥ f ∥ α LP ∥ f ∥^1 q− α
with
1 r =^
α p +^
1 − α q ,^0 ≤^ α^ ≤^ 1.
Theorem 5.4. LP^ es un espacio vectorial y ∥ · ∥LP es una norma para todo 1 ≤ p ≤ ∞.
Proof Los casos p = 1 y p = ∞ son triviales.
Supongamos que 1 < p < ∞ y f , g ∈ LP(Ω). Se tiene | f (x) + g(x)|P^ ≤ (| f (x)| + |g(x)|)P ≤ 2 P(| f (x)|P^ + |g(x)|P).
On the other hand, | f + g|P−^1 ∈ LP′ (Ω). In fact,
The Lp^ spaces 108
Z Ω^ |^ f^ +^ g|
(P− 1 )P′ =^ Z
Ω^ |^ f^ +^ g|^
PP′^ P′ =^ Z
Ω^ |^ f^ +^ g|
P < ∞.
∥ f + g∥PLP =
Z Ω^ |^ f^ +^ g|P−^1 |^ f^ +^ g| ≤
Z | f + g|P−^1 | f | +
Z | f + g|P−^1 |g| ≤ ∥| f + g|P−^1 ∥LP′ ∥ f ∥LP + ∥| f + g|P−^1 ∥LP ∥g∥LP = ∥ f + g∥P L/P P′∥ f ∥LP + ∥ f + g∥P L/P P′∥g∥LP = ∥ f + g∥P−^1 ∥ f ∥LP + ∥ f + g∥P LP− 1 ∥g∥LP = ∥ f + g∥P−^1 ∥ f ∥LP + ∥g∥PLP^ ∥ f + g∥PLP ≤ ∥ f + g∥P−^1 ∥ f ∥LP + ∥g∥PLP^ ⇒ ∥ f + g∥LP ≤ ∥ f ∥LP + ∥g∥LP. ■ Theorem 5.5. LP^ es un espacio de Banach pora 1 ≤ p ≤ ∞. Proof K ∈ N Supongamos queexiste n p = ∞. Sea ( fn) una sucesi ´on de Cauchy en L∞. Entonces, dado k ∈^ N^ tal que ∥ fm − fn∥L∞ < (^) K^1 , m, n > nK. As´ı existe un conjunto medible EK con m(EK) = 0 tal que
| fm(x) − fn(x)| ≤ (^) K^1 , x ∈ Ω\EK. Por otra parte, sabemos que: Si E = [^ EK ⇒ m(E) = 0. As´ı | fm(x) − fn(x)| ≤ (^) K^1 , x ∈ Ω\E y m, n ≥ NK. As´ı ( fn(x)) es de Cauchy para todo x ∈ Ω\E. Como ( fn(x)) ⊂ R y R es completo, se tiene que nlim→∞ fn(x) =^ f^ (x),^ x^ ∈^ Ω\E^ y^ |^ f^ (x)| ≤^ c,^ x^ ∈^ Ω\E.
The Lp^ spaces 110
Por otra parte, para m, n ≥ 2 se tiene que | fm(x) − fn(x)| ≤ | fm(x) − fm− 1 (x)| + | fm− 1 (x) − fm− 2 (x)| + · · · + | fn+ 1 (x) − fn(x)| < gm(x) − gn− 1 (x) ≤ g(x) − gn− 1 (x) (5.5) Resulta que ( fn) es de Cauchy, y por consiguiente tiene un l´ımite designado por f (x) a.e | fn(x) − f (x)|P^ < gP(x) de donde resulta queLebesgue resulta que f ∈ LP(Ω) y por el Teorema de la Convergencia Dominada de
nlim→∞ ∥^ fn^ −^ f^ ∥LP^ =^ 0. ■
Theorem 5.6. Entonces existe una subsucesi´ Sean ( fn) una sucesi´on ( f on en LP(Ω) y f ∈ LP(Ω), tales que ∥ fn − f ∥LP → 0. nk )^ tal que: a) lim fnk (x) = f (x), a.e b) | fnk (x)| ≤ h(x), a.e en Ω con h ∈ LP(Ω), ∀ k ≥ 1. Demostraci ´on El resultado es trivial si p = ∞. De la demostraci ´on del Teorema anterior resulta que existe una subsucesi ´on ( fnk ) de ( fn) tal que
∥ fnk+ 1 − fnk ∥LP ≤ (^21) k , k ≥ 1
gn(x) = k ∑=^ n 1 | fnk+ 1 (x) − fnk (x)|, a.e on Ω gn ∈ LP, ∥gn∥LP ≤ 1
se tiene por otro lado que gn+ 1 ≤ gn, a.e. y
nlim→∞ gPn^ (x) =^ gP(x),^ ∀^ x^ ∈^ Ω\E. From the Monotone Convergence Theorem, we have that g ∈ LP(Ω).
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On the other hand, for m, n ≥ 2 with m > n we consider: | fm(x) − fn(x)| ≤ | fm(x) − fm− 1 (x)| + | fm− 1 (x) − fm− 2 (x)| + · · · + | fn+ 1 (x) − fn(x)| < gm(x) − gn− 1 (x). De aqu´l´ımite que designaremos porı resulta que { fn}∞ n= (^1) fes una sucesi ´ (x), a.e y on de Cauchy, y por consiguiente tiene un
| fn(x) − f (x)|P^ ≤ gP(x), lo cual implica que f ∈ LP(Ω). Luego, aplicando el Teorema de la Convergencia Dominada de Lebesgue resulta que nlim→∞ ∥^ fn^ −^ f^ ∥LP^ =^ 0. In this case h(x) = g(x) − | f (x)|, y fk = fnk. ■ Problems
- Prove that f (x) = √ (^31) x. (a) Belongs to L[0, 8] but (b) Does not belong to L^3 [0, 8].
- If f ∈ L^2 (Ω) and g ∈ L^2 (Ω), prove that f .g ∈ L(Ω).
- Prove that L^2 (Ω) ⊂ L(Ω), if m(Ω) < ∞.
- Prove Schwarz’s inequality Z (^) b a^ f^ (x)g(x)dx^ ≤
Z (^) b a^ |^ f^ (x)|
2 1/2^ Z^ b a^ |g(x)|
(^2) dx^ 1/ where f ∈ L^2 , g ∈ L^2.
Bibliography
[Belinchon, 2008] Belinchon, J. A. (2008). Apuntes de Teoria de la Medida. [Gatica, 1997] Gatica, J. A. (1997). Introduccion a la integral de lebesgue en la recta. [Miguel De Guzman y Baldomero Rubio, 1979] Miguel De Guzman y Baldomero Ru-bio. (1979). Integracion: Teoria y Tecnica. Editorial Alhambra, S.A.
[Spiegel, 1990] Spiegel, M. R. (1990).sure and Integral. Schaum’s Outline Series. Theory and Problems of Real Variables: Lebesgue Mea-
[Howard Wilcox and David Meyer, 1978] Howard Wilcox and David Meyer. (1978). An Introduction to Lebesgue Integral and Fourier Seriesington, New York, 1978.. Robert E. Kriger Publishing Hunt-
