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$L^p$ Spaces

For 1p<1 \leq p < \infty, define

Lp(μ)={f:XR measurable:Xfpdμ<}L^p(\mu) = \left\{f : X \to \mathbb{R} \text{ measurable} : \int_X |f|^p\, d\mu < \infty\right\}

with the norm fp=(fpdμ)1/p\|f\|_p = \left(\int |f|^p\, d\mu\right)^{1/p}.

For p=p = \infty, define L(μ)={f:XR measurable:ess supf<}L^\infty(\mu) = \{f : X \to \mathbb{R} \text{ measurable} : \text{ess sup}|f| < \infty\} where f=esssupf\|f\|_\infty = \mathrm{ess\,sup}|f|.

Remark. Elements of LpL^p are equivalence classes of functions equal a.e. The norm p\|\cdot\|_p is well-defined on equivalence classes.

Theorem 7.1 (Holder’s Inequality). Let 1p,q1 \leq p, q \leq \infty with 1/p+1/q=11/p + 1/q = 1. If fLp(μ)f \in L^p(\mu) and gLq(μ)g \in L^q(\mu), then fgL1(μ)fg \in L^1(\mu) and

fg1fpgq\|fg\|_1 \leq \|f\|_p \cdot \|g\|_q

Proof sketch. Use Young’s inequality: abap/p+bq/qab \leq a^p/p + b^q/q for a,b0a, b \geq 0. Set a=f/fpa = |f|/\|f\|_p and b=g/gqb = |g|/\|g\|_q and integrate. \blacksquare

Special case (p=q=2p = q = 2): This reduces to the Cauchy-Schwarz inequality: fg1f2g2\|fg\|_1 \leq \|f\|_2 \|g\|_2.

Theorem 7.2 (Minkowski’s Inequality). For 1p1 \leq p \leq \infty and f,gLp(μ)f, g \in L^p(\mu):

f+gpfp+gp\|f + g\|_p \leq \|f\|_p + \|g\|_p

Proof sketch (for 1<p<1 < p < \infty). Write f+gp=f+gf+gp1|f + g|^p = |f + g| \cdot |f + g|^{p-1}. Apply Holder’s inequality with conjugate exponents pp and q=p/(p1)q = p/(p-1):

f+gpfpf+gpp/q+gpf+gpp/q\int |f + g|^p \leq \|f\|_p \|f + g\|_p^{p/q} + \|g\|_p \|f + g\|_p^{p/q}

Divide both sides by f+gpp/q\|f + g\|_p^{p/q}. \blacksquare

Theorem 7.3. (Lp(μ),p)(L^p(\mu), \|\cdot\|_p) is a Banach space for every 1p1 \leq p \leq \infty.

Proof sketch. Let {fn}\{f_n\} be a Cauchy sequence in LpL^p. Extract a subsequence fnkf_{n_k} with fnk+1fnkp<2k\|f_{n_{k+1}} - f_{n_k}\|_p < 2^{-k}. Define g=fn1+k=1fnk+1fnkg = |f_{n_1}| + \sum_{k=1}^{\infty} |f_{n_{k+1}} - f_{n_k}|. By the triangle inequality, gpfn1p+1\|g\|_p \leq \|f_{n_1}\|_p + 1. So gLpg \in L^p, hence g<g < \infty a.e., meaning fnkf_{n_k} converges a.e. to some ff. Show fLpf \in L^p and fnff_n \to f in LpL^p-norm. \blacksquare

Theorem 7.4. L2(μ)L^2(\mu) is a Hilbert space with inner product f,g=fgdμ\langle f, g \rangle = \int f\overline{g}\, d\mu.

Proposition 7.5. If μ\mu is a finite measure and 1p<q1 \leq p < q \leq \infty, then Lq(μ)Lp(μ)L^q(\mu) \subseteq L^p(\mu). In particular, L(μ)L2(μ)L1(μ)L^\infty(\mu) \subseteq L^2(\mu) \subseteq L^1(\mu).

Proof. Apply Holder’s inequality with q/pq/p and its conjugate. \blacksquare

Theorem 7.6 (Riesz Representation for LpL^p). For 1<p<1 < p < \infty, the dual space (Lp)(L^p)^* is isometrically isomorphic to LqL^q, where 1/p+1/q=11/p + 1/q = 1. The pairing is:

f,g=fgdμ,fLp,gLq\langle f, g \rangle = \int f g\, d\mu, \quad f \in L^p, g \in L^q

This also holds for p=1p = 1 provided μ\mu is σ\sigma-finite (the dual of L1L^1 is LL^\infty). For p=p = \infty, the dual is strictly larger than L1L^1 (except for finite-dimensional spaces).

Proposition 7.7. For 1<p<1 < p < \infty, LpL^p spaces are uniformly convex: for any ε>0\varepsilon > 0, there exists δ>0\delta > 0 such that if fp=gp=1\|f\|_p = \|g\|_p = 1 and fgpε\|f - g\|_p \geq \varepsilon, then (f+g)/2p1δ\|(f + g)/2\|_p \leq 1 - \delta.

Uniform convexity implies reflexivity for 1<p<1 < p < \infty 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)L^p(\mu) for 1p<1 \leq p < \infty. For the Lebesgue measure on Rn\mathbb{R}^n, the following are also dense:

  1. Step functions (finite linear combinations of characteristic functions of rectangles)
  2. Continuous functions with compact support Cc(Rn)C_c(\mathbb{R}^n)
  3. Smooth functions with compact support Cc(Rn)C_c^\infty(\mathbb{R}^n)

Proposition 7.9. If fnff_n \to f in LpL^p, then there exists a subsequence fnkf_{n_k} that converges pointwise a.e. to ff. The converse is false: pointwise convergence a.e. does not imply LpL^p convergence (counterexample: fn=nχ[0,1/n]f_n = n\chi_{[0, 1/n]} on [0,1][0, 1] converges pointwise to 00 but fn1=1\|f_n\|_1 = 1 for all nn).

Theorem 7.10 (Dominated Convergence in LpL^p). If fnff_n \to f a.e. and there exists gLpg \in L^p such that fng|f_n| \leq g a.e. for all nn, then fnff_n \to f in LpL^p.

7.10 Worked Example: LpL^p Norm Behaviour

Section titled “7.10 Worked Example: LpL^pLp Norm Behaviour”

Problem. Let f(x)=x1/2f(x) = x^{-1/2} on (0,1)(0, 1) with Lebesgue measure. For which pp does fLp(0,1)f \in L^p(0, 1)?

Solution

Compute 01f(x)pdx=01xp/2dx\int_0^1 |f(x)|^p dx = \int_0^1 x^{-p/2} dx. This integral converges iff p/2>1-p/2 > -1, i.e., p<2p < 2. So fLp(0,1)f \in L^p(0, 1) precisely for 1p<21 \leq p < 2. Note fL2(0,1)f \notin L^2(0, 1) (the integral diverges logarithmically at the boundary).

\blacksquare

Problem. Show that if fLpLqf \in L^p \cap L^q with 1p<q1 \leq p < q \leq \infty, then fLrf \in L^r for all r[p,q]r \in [p, q].

Solution

Write r=θp+(1θ)qr = \theta p + (1-\theta)q with θ[0,1]\theta \in [0, 1], so 1=θp/r+(1θ)q/r1 = \theta p/r + (1-\theta)q/r. Apply Holder’s inequality with exponents r/(θp)r/(\theta p) and r/((1θ)q)r/((1-\theta)q):

fr=fθpf(1θ)q(fp)θr/p(fq)(1θ)r/q\int |f|^r = \int |f|^{\theta p} |f|^{(1-\theta)q} \leq \left(\int |f|^p\right)^{\theta r/p} \left(\int |f|^q\right)^{(1-\theta) r/q}

Therefore frfpθfq1θ\|f\|_r \leq \|f\|_p^\theta \|f\|_q^{1-\theta}, establishing both that fLrf \in L^r and a quantitative interpolation inequality.

\blacksquare

Definition. The weak LpL^p space Lp,(μ)L^{p,\infty}(\mu) consists of measurable functions for which

[f]p,=supt>0tμ({x:f(x)>t})1/p<[f]_{p,\infty} = \sup_{t > 0} t\, \mu(\{x : |f(x)| > t\})^{1/p} < \infty

Weak LpL^p spaces are larger than LpL^p: LpLp,L^p \subseteq L^{p,\infty} with fp,fp\|f\|_{p,\infty} \leq \|f\|_p.

Example. The function f(x)=1/x1/pf(x) = 1/|x|^{1/p} on R\mathbb{R} is in Lp,L^{p,\infty} but not in LpL^p (the singularity is just barely non-integrable in the LpL^p sense).

Lorentz spaces Lp,q(μ)L^{p,q}(\mu) refine the LpL^p scale, with Lp,p=LpL^{p,p} = L^p and Lp,L^{p,\infty} being weak LpL^p. They are important in interpolation theory and harmonic analysis.

7.13 Worked Example: LpL^p on a Finite Measure Space

Section titled “7.13 Worked Example: LpL^pLp on a Finite Measure Space”

Problem. Show that if μ(X)<\mu(X) < \infty, then limpfp=f\lim_{p \to \infty} \|f\|_p = \|f\|_\infty for fL(μ)f \in L^\infty(\mu).

Solution

Let M=fM = \|f\|_\infty. For any ε>0\varepsilon > 0, the set A={x:f(x)>Mε}A = \{x : |f(x)| > M - \varepsilon\} has μ(A)>0\mu(A) > 0. Then:

fp(A(Mε)pdμ)1/p=(Mε)μ(A)1/p\|f\|_p \geq \left(\int_A (M - \varepsilon)^p d\mu\right)^{1/p} = (M - \varepsilon)\,\mu(A)^{1/p}

As pp \to \infty, μ(A)1/p1\mu(A)^{1/p} \to 1, so lim infpfpMε\liminf_{p\to\infty} \|f\|_p \geq M - \varepsilon. Since ε\varepsilon is arbitrary, lim inffpM\liminf \|f\|_p \geq M.

Conversely, fpMμ(X)1/p\|f\|_p \leq M\,\mu(X)^{1/p}, so lim suppfpM\limsup_{p\to\infty} \|f\|_p \leq M. Therefore limpfp=M=f\lim_{p\to\infty} \|f\|_p = M = \|f\|_\infty.

\blacksquare