For 1≤p<∞, define
Lp(μ)={f:X→R measurable:∫X∣f∣pdμ<∞}
with the norm ∥f∥p=(∫∣f∣pdμ)1/p.
For p=∞, define L∞(μ)={f:X→R measurable:ess sup∣f∣<∞} where ∥f∥∞=esssup∣f∣.
Remark. Elements of Lp are equivalence classes of functions equal a.e. The norm ∥⋅∥p is well-defined on equivalence classes.
Theorem 7.1 (Holder’s Inequality). Let 1≤p,q≤∞ with 1/p+1/q=1. If f∈Lp(μ) and g∈Lq(μ), then fg∈L1(μ) and
∥fg∥1≤∥f∥p⋅∥g∥q
Proof sketch. Use Young’s inequality: ab≤ap/p+bq/q for a,b≥0. Set a=∣f∣/∥f∥p and b=∣g∣/∥g∥q and integrate. ■
Special case (p=q=2): This reduces to the Cauchy-Schwarz inequality: ∥fg∥1≤∥f∥2∥g∥2.
Theorem 7.2 (Minkowski’s Inequality). For 1≤p≤∞ and f,g∈Lp(μ):
∥f+g∥p≤∥f∥p+∥g∥p
Proof sketch (for 1<p<∞). Write ∣f+g∣p=∣f+g∣⋅∣f+g∣p−1. Apply Holder’s inequality with conjugate exponents p and q=p/(p−1):
∫∣f+g∣p≤∥f∥p∥f+g∥pp/q+∥g∥p∥f+g∥pp/q
Divide both sides by ∥f+g∥pp/q. ■
Theorem 7.3. (Lp(μ),∥⋅∥p) is a Banach space for every 1≤p≤∞.
Proof sketch. Let {fn} be a Cauchy sequence in Lp. Extract a subsequence fnk with ∥fnk+1−fnk∥p<2−k. Define g=∣fn1∣+∑k=1∞∣fnk+1−fnk∣. By the triangle inequality, ∥g∥p≤∥fn1∥p+1. So g∈Lp, hence g<∞ a.e., meaning fnk converges a.e. to some f. Show f∈Lp and fn→f in Lp-norm. ■
Theorem 7.4. L2(μ) is a Hilbert space with inner product ⟨f,g⟩=∫fgdμ.
Proposition 7.5. If μ is a finite measure and 1≤p<q≤∞, then Lq(μ)⊆Lp(μ). In particular, L∞(μ)⊆L2(μ)⊆L1(μ).
Proof. Apply Holder’s inequality with q/p and its conjugate. ■
Theorem 7.6 (Riesz Representation for Lp). For 1<p<∞, the dual space (Lp)∗ is isometrically isomorphic to Lq, where 1/p+1/q=1. The pairing is:
⟨f,g⟩=∫fgdμ,f∈Lp,g∈Lq
This also holds for p=1 provided μ is σ-finite (the dual of L1 is L∞). For p=∞, the dual is strictly larger than L1 (except for finite-dimensional spaces).
Proposition 7.7. For 1<p<∞, Lp spaces are uniformly convex: for any ε>0, there exists δ>0 such that if ∥f∥p=∥g∥p=1 and ∥f−g∥p≥ε, then ∥(f+g)/2∥p≤1−δ.
Uniform convexity implies reflexivity for 1<p<∞ and guarantees the existence and uniqueness of best approximations in closed convex subspaces.
Proposition 7.8 (Density of Simple Functions). Simple functions are dense in Lp(μ) for 1≤p<∞. For the Lebesgue measure on Rn, the following are also dense:
- Step functions (finite linear combinations of characteristic functions of rectangles)
- Continuous functions with compact support Cc(Rn)
- Smooth functions with compact support Cc∞(Rn)
Proposition 7.9. If fn→f in Lp, then there exists a subsequence fnk that converges pointwise a.e. to f. The converse is false: pointwise convergence a.e. does not imply Lp convergence (counterexample: fn=nχ[0,1/n] on [0,1] converges pointwise to 0 but ∥fn∥1=1 for all n).
Theorem 7.10 (Dominated Convergence in Lp). If fn→f a.e. and there exists g∈Lp such that ∣fn∣≤g a.e. for all n, then fn→f in Lp.
Problem. Let f(x)=x−1/2 on (0,1) with Lebesgue measure. For which p does f∈Lp(0,1)?
Solution
Compute ∫01∣f(x)∣pdx=∫01x−p/2dx. This integral converges iff −p/2>−1, i.e., p<2. So f∈Lp(0,1) precisely for 1≤p<2. Note f∈/L2(0,1) (the integral diverges logarithmically at the boundary).
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Problem. Show that if f∈Lp∩Lq with 1≤p<q≤∞, then f∈Lr for all r∈[p,q].
Solution
Write r=θp+(1−θ)q with θ∈[0,1], so 1=θp/r+(1−θ)q/r. Apply Holder’s inequality with exponents r/(θp) and r/((1−θ)q):
∫∣f∣r=∫∣f∣θp∣f∣(1−θ)q≤(∫∣f∣p)θr/p(∫∣f∣q)(1−θ)r/q
Therefore ∥f∥r≤∥f∥pθ∥f∥q1−θ, establishing both that f∈Lr and a quantitative interpolation inequality.
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Definition. The weak Lp space Lp,∞(μ) consists of measurable functions for which
[f]p,∞=supt>0tμ({x:∣f(x)∣>t})1/p<∞
Weak Lp spaces are larger than Lp: Lp⊆Lp,∞ with ∥f∥p,∞≤∥f∥p.
Example. The function f(x)=1/∣x∣1/p on R is in Lp,∞ but not in Lp (the singularity is just barely non-integrable in the Lp sense).
Lorentz spaces Lp,q(μ) refine the Lp scale, with Lp,p=Lp and Lp,∞ being weak Lp. They are important in interpolation theory and harmonic analysis.
Problem. Show that if μ(X)<∞, then limp→∞∥f∥p=∥f∥∞ for f∈L∞(μ).
Solution
Let M=∥f∥∞. For any ε>0, the set A={x:∣f(x)∣>M−ε} has μ(A)>0. Then:
∥f∥p≥(∫A(M−ε)pdμ)1/p=(M−ε)μ(A)1/p
As p→∞, μ(A)1/p→1, so liminfp→∞∥f∥p≥M−ε. Since ε is arbitrary, liminf∥f∥p≥M.
Conversely, ∥f∥p≤Mμ(X)1/p, so limsupp→∞∥f∥p≤M. Therefore limp→∞∥f∥p=M=∥f∥∞.
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