Book 4A

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Course 4Book 4A: Function Spaces and Sobolev SpacesChapter 12

Interpolation and Fractional Sobolev Spaces

Riesz–Thorin and the three lines lemma, Young and Hausdorff–Young, fractional smoothness Hˢ and Wˢ,ᵖ, traces on the boundary, Gagliardo–Nirenberg interpolation, and the fractional Laplacian.

39 min read · Updated Oct 24, 2026

Read with Grafakos, Classical Fourier Analysis, §1.3 (Riesz–Thorin and Marcinkiewicz) and §2.2 (Hausdorff–Young), or Stein and Shakarchi, Functional Analysis, chapter 2 (L^p spaces and interpolation). For fractional Sobolev spaces, Di Nezza, Palatucci and Valdinoci, "Hitchhiker's guide to the fractional Sobolev spaces" (Bulletin des Sciences Mathématiques, 2012), sections 2–3 and 6, is short and complete; Evans §5.5 treats traces of W1,pW^{1,p}.

In this chapter · 8 sections
  1. 12.1Jumps of all sizes
  2. 12.2The Riesz–Thorin theorem
  3. 12.3Fractional Sobolev spaces
  4. 12.4Interpolation inequalities
  5. 12.5Traces
  6. 12.6The fractional Laplacian
  7. 12.7History
  8. 12.8Exercises
  9. ↗Free resources

The Sobolev spaces of 4A.10 Sobolev Spaces and 4A.11 Sobolev Embeddings and Critical Exponents count derivatives in whole numbers: W1,pW^{1,p}, W2,pW^{2,p}, HkH^k. Three questions force fractional smoothness on us.

  • Boundary values. A function in H1(Ω)H^1(\Omega) has a trace on ∂Ω\partial\Omega, but the trace is not in H1(∂Ω)H^1(\partial\Omega), nor merely in L2(∂Ω)L^2(\partial\Omega): it has exactly half a derivative, H1/2H^{1/2}. The Dirichlet problem with boundary data cannot be stated sharply without this.
  • Nonlocal operators. Diffusion driven by jumps of all sizes, rather than by Brownian motion, is governed by the fractional Laplacian (−Δ)s(-\Delta)^s, 0<s<10 < s < 1, whose natural energy space is HsH^s.
  • Interpolation. If an operator is bounded on L1L^1 and on L2L^2, is it bounded on LpL^p in between? If a function is in L2L^2 and in H2H^2, how much smoothness does it have "in between"? Interpolation theorems answer both with a convexity statement: norms in between are controlled by the norms at the ends, with exponents.

This chapter proves the Riesz–Thorin interpolation theorem, with the Hausdorff–Young and Young inequalities as corollaries; defines HsH^s by the Fourier transform and Ws,pW^{s,p} by difference quotients, and proves that they agree for p=2p = 2; proves the trace theorem Hs(Rn)→Hs−1/2(Rn−1)H^s(\mathbb{R}^n) \to H^{s - 1/2}(\mathbb{R}^{n - 1}); proves the interpolation inequalities of Gagliardo–Nirenberg type that the PDE of Books 6 and 11 use constantly; and introduces the fractional Laplacian. The Marcinkiewicz theorem, the "real" counterpart of Riesz–Thorin, is proved in 6B.8 Lᵖ Theory and Strong Solutions, where singular integrals need it.

By the end of this chapter you will be able to:

  • prove the three lines lemma and the Riesz–Thorin theorem, and deduce the Hausdorff–Young and Young inequalities;
  • define Hs(Rn)H^s(\mathbb{R}^n) for real ss and Ws,pW^{s,p} for 0<s<10 < s < 1, and prove the equivalence of the Fourier and difference-quotient definitions when p=2p = 2;
  • prove the interpolation inequality ∥u∥Hs≤∥u∥Hs01−θ∥u∥Hs1θ\|u\|_{H^s} \leq \|u\|_{H^{s_0}}^{1 - \theta}\|u\|_{H^{s_1}}^\theta and the one-dimensional inequality ∥u∥∞2≤∥u∥2∥u′∥2\|u\|_\infty^2 \leq \|u\|_2\|u'\|_2, and state Gagliardo–Nirenberg;
  • prove the trace theorem and explain why the trace space of H1H^1 is H1/2H^{1/2};
  • define the fractional Laplacian by its symbol and as a singular integral, and explain its probabilistic meaning.

Jumps of all sizes

In the world Model Lévy flights, and a cautionary tale

A Brownian particle moves by many small independent steps, and its density obeys the heat equation (6A.3 The Heat Equation on ℝⁿ). If the steps have a heavy-tailed distribution, with probability of a step longer than rr decaying like r−2sr^{-2s} for some 0<s<10 < s < 1, occasional long jumps dominate and the density obeys ∂tu=−(−Δ)su\partial_tu = -(-\Delta)^su: anomalous diffusion, or a Lévy flight. Such models are used for light in disordered media, turbulent transport and price changes in finance. In 1996 Gandhimohan Viswanathan and coauthors reported in Nature that wandering albatrosses forage by Lévy flights, and the idea that Lévy flights are an optimal search strategy became influential in ecology. In 2007 Andrew Edwards and coauthors, also in Nature, reanalysed better data and found that the albatross evidence did not support Lévy flights after all: some recorded "flights" had been time the birds spent sitting on the water.

Where the picture breaks

The mathematics is not at fault in the albatross story: the fractional heat equation correctly describes particles that jump with power-law tails. What failed was the identification of the data with that model. Distinguishing a power-law tail from a mixture of ordinary exponentials needs long records and careful statistics, and is a standing difficulty in applications of heavy-tailed models.

The Riesz–Thorin theorem

Complex interpolation rests on one lemma about holomorphic functions. Let S={z:0≤Re⁡z≤1}S = \{z : 0 \leq \operatorname{Re}z \leq 1\}, a closed strip.

Lemma 12.1 Three lines lemma (Hadamard)

Let FF be continuous and bounded on SS and holomorphic in its interior, with ∣F(it)∣≤M0|F(it)| \leq M_0 and ∣F(1+it)∣≤M1|F(1 + it)| \leq M_1 for all real tt. Then ∣F(θ+it)∣≤M01−θM1θ|F(\theta + it)| \leq M_0^{1 - \theta}M_1^\theta for 0≤θ≤10 \leq \theta \leq 1.

Proof. Assume M0,M1>0M_0, M_1 > 0 (otherwise replace them by Mi+δM_i + \delta and let δ→0\delta \to 0). The function G(z)=F(z)M0z−1M1−zG(z) = F(z)M_0^{z - 1}M_1^{-z} is bounded on SS, holomorphic inside, and ∣G∣≤1|G| \leq 1 on both boundary lines, since ∣M0z−1M1−z∣=M0Re⁡z−1M1−Re⁡z|M_0^{z - 1}M_1^{-z}| = M_0^{\operatorname{Re}z - 1}M_1^{-\operatorname{Re}z}. For ε>0\varepsilon > 0, Gε(z)=G(z)eεz2G_\varepsilon(z) = G(z)e^{\varepsilon z^2} satisfies ∣eεz2∣=eε(x2−y2)≤eεe−εy2|e^{\varepsilon z^2}| = e^{\varepsilon(x^2 - y^2)} \leq e^\varepsilon e^{-\varepsilon y^2} for z=x+iy∈Sz = x + iy \in S; so ∣Gε∣≤eε|G_\varepsilon| \leq e^\varepsilon on the boundary and Gε→0G_\varepsilon \to 0 as ∣y∣→∞|y| \to \infty in SS. By the maximum modulus principle (5A.2 Cauchy’s Theorem and Its Consequences) on a tall rectangle [0,1]×[−Y,Y][0, 1]\times[-Y, Y], ∣Gε∣≤eε|G_\varepsilon| \leq e^\varepsilon inside, for every YY large. Let ε→0\varepsilon \to 0: ∣G∣≤1|G| \leq 1 on SS, which is the claim.

The lemma says that θ↦log⁡sup⁡t∣F(θ+it)∣\theta \mapsto \log\sup_t|F(\theta + it)| is a convex function: the logarithm of a maximum of a holomorphic function is convex across a strip, a cousin of the convexity of log⁡\log-moments (3A.7 Lᵖ Spaces and Jensen’s Inequality). The factor eεz2e^{\varepsilon z^2} handles the unbounded strip, and some growth condition is essential: F(z)=exp⁡(exp⁡(−iπ(z−12)))F(z) = \exp\big(\exp(-i\pi(z - \frac12))\big) has modulus 11 on both boundary lines (where Re⁡exp⁡(−iπ(z−12))=0\operatorname{Re}\exp(-i\pi(z - \frac12)) = 0) but modulus exp⁡(eπy)\exp(e^{\pi y}) at z=12+iyz = \frac12 + iy.

Theorem 12.2 Riesz–Thorin interpolation theorem

Let 1≤p0,p1,q0,q1≤∞1 \leq p_0, p_1, q_0, q_1 \leq \infty, and let TT be a linear operator defined on simple functions (on σ\sigma-finite measure spaces), with

∥Tf∥q0≤M0∥f∥p0,∥Tf∥q1≤M1∥f∥p1.\|Tf\|_{q_0} \leq M_0\|f\|_{p_0}, \qquad \|Tf\|_{q_1} \leq M_1\|f\|_{p_1}.

For 0<θ<10 < \theta < 1 define pθp_\theta, qθq_\theta by 1pθ=1−θp0+θp1\frac{1}{p_\theta} = \frac{1 - \theta}{p_0} + \frac{\theta}{p_1} and 1qθ=1−θq0+θq1\frac{1}{q_\theta} = \frac{1 - \theta}{q_0} + \frac{\theta}{q_1}. Then ∥Tf∥qθ≤M01−θM1θ∥f∥pθ\|Tf\|_{q_\theta} \leq M_0^{1 - \theta}M_1^\theta\|f\|_{p_\theta}, and TT extends to a bounded operator Lpθ→LqθL^{p_\theta} \to L^{q_\theta} when pθ<∞p_\theta < \infty.

Proof. (Complex scalars.) By duality (4A.4 Hahn–Banach and Duality), ∥Tf∥qθ=sup⁡∣∫(Tf)g∣\|Tf\|_{q_\theta} = \sup\big|\int(Tf)g\big| over simple gg with ∥g∥qθ′=1\|g\|_{q_\theta'} = 1, where q′q' is the conjugate exponent. Fix simple ff with ∥f∥pθ=1\|f\|_{p_\theta} = 1 and such gg. Write f=∣f∣eiαf = |f|e^{i\alpha}, g=∣g∣eiβg = |g|e^{i\beta}, and for z∈Sz \in S define

fz=∣f∣pθ(1−zp0+zp1)eiα,gz=∣g∣qθ′(1−zq0′+zq1′)eiβ,f_z = |f|^{p_\theta\big(\frac{1 - z}{p_0} + \frac{z}{p_1}\big)}e^{i\alpha}, \qquad g_z = |g|^{q_\theta'\big(\frac{1 - z}{q_0'} + \frac{z}{q_1'}\big)}e^{i\beta},

so that fθ=ff_\theta = f and gθ=gg_\theta = g. Since ff and gg are simple, F(z)=∫(Tfz)gzF(z) = \int(Tf_z)g_z is a finite sum of exponentials ckeakzc_ke^{a_kz}: entire and bounded on SS. On Re⁡z=0\operatorname{Re}z = 0, ∣fit∣=∣f∣pθ/p0|f_{it}| = |f|^{p_\theta/p_0}, so ∥fit∥p0p0=∫∣f∣pθ=1\|f_{it}\|_{p_0}^{p_0} = \int|f|^{p_\theta} = 1, and likewise ∥git∥q0′=1\|g_{it}\|_{q_0'} = 1; hence ∣F(it)∣≤∥Tfit∥q0∥git∥q0′≤M0|F(it)| \leq \|Tf_{it}\|_{q_0}\|g_{it}\|_{q_0'} \leq M_0. Similarly ∣F(1+it)∣≤M1|F(1 + it)| \leq M_1. The three lines lemma gives ∣∫(Tf)g∣=∣F(θ)∣≤M01−θM1θ|\int(Tf)g| = |F(\theta)| \leq M_0^{1 - \theta}M_1^\theta. (When some exponent is infinite the corresponding factor is read as 11; Exercise 12.10.) Simple functions are dense in LpθL^{p_\theta} for pθ<∞p_\theta < \infty, so TT extends.

The theorem is a convexity statement: log⁡∥T∥L1/a→L1/b\log\|T\|_{L^{1/a}\to L^{1/b}} is a convex function of (a,b)=(1p,1q)(a, b) = (\frac1p, \frac1q) on the square [0,1]2[0, 1]^2 (the Riesz diagram). Marcel Riesz proved it in 1926 for real scalars with the restriction p≤qp \leq q, by a variational argument; Olof Thorin's 1939 proof with complex methods removed the restriction and introduced the three lines idea that later became the general theory of complex interpolation of Banach spaces (Calderón, 1964).

Two corollaries.

Corollary 12.3 Hausdorff–Young inequality

For 1≤p≤21 \leq p \leq 2 and 1p+1p′=1\frac1p + \frac1{p'} = 1, the Fourier transform satisfies ∥f^∥p′≤∥f∥p\|\hat f\|_{p'} \leq \|f\|_p.

Proof. ∥f^∥∞≤∥f∥1\|\hat f\|_\infty \leq \|f\|_1 trivially, and ∥f^∥2=∥f∥2\|\hat f\|_2 = \|f\|_2 by Plancherel (4A.6 The Fourier Transform). Interpolate with p0=1p_0 = 1, q0=∞q_0 = \infty, p1=q1=2p_1 = q_1 = 2: 1pθ=1−θ2\frac1{p_\theta} = 1 - \frac\theta2 and 1qθ=θ2=1−1pθ\frac1{q_\theta} = \frac\theta2 = 1 - \frac1{p_\theta}.

The constant 11 is not sharp for 1<p<21 < p < 2: Babenko (1961, for even conjugates) and Beckner (1975) proved that the best constant is (p1/p/p′1/p′)n/2(p^{1/p}/p'^{1/p'})^{n/2}, attained by Gaussians, a theme that runs from the uncertainty principle (4A.6 The Fourier Transform) to entropy: differentiating Beckner's sharp inequality at p=2p = 2 gives the entropic uncertainty principle conjectured by Hirschman, a relative of the logarithmic Sobolev inequality of 6A.12 Entropy, Information and Diffusion. For p>2p > 2 there is no inequality at all: the Fourier transform of an LpL^p function need not be a function (Exercise 12.11).

Corollary 12.4 Young's convolution inequality

If f∈Lp(Rn)f \in L^p(\mathbb{R}^n), g∈Lr(Rn)g \in L^r(\mathbb{R}^n) and 1+1q=1p+1r1 + \frac1q = \frac1p + \frac1r with 1≤p,q,r≤∞1 \leq p, q, r \leq \infty, then ∥f∗g∥q≤∥f∥p∥g∥r\|f*g\|_q \leq \|f\|_p\|g\|_r.

Proof. Fix g∈Lrg \in L^r and let Tf=f∗gTf = f*g. By Minkowski's integral inequality, ∥f∗g∥r≤∥f∥1∥g∥r\|f*g\|_r \leq \|f\|_1\|g\|_r; by Hölder, ∥f∗g∥∞≤∥f∥r′∥g∥r\|f*g\|_\infty \leq \|f\|_{r'}\|g\|_r. Interpolate between (p0,q0)=(1,r)(p_0, q_0) = (1, r) and (p1,q1)=(r′,∞)(p_1, q_1) = (r', \infty): 1p=1−θr\frac1p = 1 - \frac\theta r, 1q=1−θr\frac1q = \frac{1 - \theta}{r}, and indeed 1p+1r−1=1r−θr=1q\frac1p + \frac1r - 1 = \frac1r - \frac\theta r = \frac1q.

Young's inequality is how the heat kernel acts on LpL^p: since ∥Φt∥r=crt−n2(1−1r)\|\Phi_t\|_r = c_rt^{-\frac n2(1 - \frac1r)} for the Gaussian kernel (6A.3 The Heat Equation on ℝⁿ), ∥etΔf∥q≤Ct−n2(1p−1q)∥f∥p\|e^{t\Delta}f\|_q \leq Ct^{-\frac n2(\frac1p - \frac1q)}\|f\|_p: the LpL^p–LqL^q smoothing estimate that drives semilinear theory (6A.8 Nonlinear Parabolic Equations).

Fractional Sobolev spaces

The Fourier definition. For real ss, let Hs(Rn)H^s(\mathbb{R}^n) be the tempered distributions uu with u^\hat u a function and

∥u∥Hs2=∫Rn(1+∣ξ∣2)s∣u^(ξ)∣2 dξ<∞\|u\|_{H^s}^2 = \int_{\mathbb{R}^n}(1 + |\xi|^2)^s|\hat u(\xi)|^2\,d\xi < \infty

(the convention of 4A.6 The Fourier Transform, with u^(ξ)=∫ue−2πix⋅ξ\hat u(\xi) = \int ue^{-2\pi ix\cdot\xi}; other conventions change ∣ξ∣2|\xi|^2 to 4π2∣ξ∣24\pi^2|\xi|^2 and alter constants only). For s=ks = k a nonnegative integer this is Hk=Wk,2H^k = W^{k,2} with an equivalent norm, since ∂αu^=(2πiξ)αu^\widehat{\partial^\alpha u} = (2\pi i\xi)^\alpha\hat u and (1+∣ξ∣2)k(1 + |\xi|^2)^k is comparable to ∑∣α∣≤k∣ξα∣2\sum_{|\alpha|\leq k}|\xi^\alpha|^2. Negative ss are allowed: H−sH^{-s} is the dual of HsH^s under the L2L^2 pairing, and δ0∈H−s\delta_0 \in H^{-s} exactly when s>n2s > \frac n2, since δ^0=1\hat\delta_0 = 1.

The difference-quotient definition. For 0<s<10 < s < 1 and 1≤p<∞1 \leq p < \infty, the Gagliardo (Slobodeckij) seminorm is

[u]Ws,pp=∫ ⁣ ⁣∫Rn×Rn∣u(x)−u(y)∣p∣x−y∣n+sp dx dy,[u]_{W^{s,p}}^p = \int\!\!\int_{\mathbb{R}^n\times\mathbb{R}^n}\frac{|u(x) - u(y)|^p}{|x - y|^{n + sp}}\,dx\,dy,

and Ws,p={u∈Lp:[u]Ws,p<∞}W^{s,p} = \{u \in L^p : [u]_{W^{s,p}} < \infty\}. It asks for a "fractional Hölder" bound in an averaged sense: ∣u(x)−u(y)∣|u(x) - u(y)| should be small compared with ∣x−y∣s|x - y|^s, on average. It makes sense on any domain Ω\Omega (integrate over Ω×Ω\Omega\times\Omega) and on manifolds, where there is no Fourier transform.

Proposition 12.5 The two definitions agree for p=2p = 2

For 0<s<10 < s < 1 there is C(n,s)>0C(n, s) > 0 with [u]Ws,22=C(n,s)∫∣ξ∣2s∣u^(ξ)∣2 dξ[u]_{W^{s,2}}^2 = C(n, s)\int|\xi|^{2s}|\hat u(\xi)|^2\,d\xi for all u∈L2u \in L^2. Hence Ws,2(Rn)=Hs(Rn)W^{s,2}(\mathbb{R}^n) = H^s(\mathbb{R}^n) with equivalent norms.

Proof. Substitute y=x+hy = x + h: [u]2=∫1∣h∣n+2s∫∣u(x+h)−u(x)∣2 dx dh[u]^2 = \int\frac{1}{|h|^{n + 2s}}\int|u(x + h) - u(x)|^2\,dx\,dh. By Plancherel, the inner integral is ∫∣e2πih⋅ξ−1∣2∣u^(ξ)∣2dξ\int|e^{2\pi ih\cdot\xi} - 1|^2|\hat u(\xi)|^2d\xi. By Tonelli,

[u]2=∫∣u^(ξ)∣2 I(ξ) dξ,I(ξ)=∫Rn∣e2πih⋅ξ−1∣2∣h∣n+2s dh.[u]^2 = \int|\hat u(\xi)|^2\,I(\xi)\,d\xi, \qquad I(\xi) = \int_{\mathbb{R}^n}\frac{|e^{2\pi ih\cdot\xi} - 1|^2}{|h|^{n + 2s}}\,dh.

II is rotation invariant, and substituting h=k/∣ξ∣h = k/|\xi| shows I(ξ)=∣ξ∣2sI(e1)I(\xi) = |\xi|^{2s}I(e_1). Finally I(e1)<∞I(e_1) < \infty: near h=0h = 0, ∣e2πih1−1∣2≤4π2∣h∣2|e^{2\pi ih_1} - 1|^2 \leq 4\pi^2|h|^2 and ∫∣h∣<1∣h∣2−n−2s<∞\int_{|h|<1}|h|^{2 - n - 2s} < \infty since s<1s < 1; near infinity the numerator is at most 44 and ∫∣h∣>1∣h∣−n−2s<∞\int_{|h|>1}|h|^{-n - 2s} < \infty since s>0s > 0.

At s=1s = 1 the constant blows up, and (1−s)[u]Ws,pp→c∥∇u∥pp(1 - s)[u]_{W^{s,p}}^p \to c\|\nabla u\|_p^p as s→1s \to 1 (Bourgain, Brezis and Mironescu, 2001): the fractional seminorms connect continuously to the classical ones after renormalisation. At s=0s = 0, s[u]Ws,pp→c∥u∥pps[u]_{W^{s,p}}^p \to c\|u\|_p^p (Maz'ya and Shaposhnikova, 2002).

Interpolation of HsH^s. The Fourier definition makes interpolation a matter of Hölder's inequality.

Proposition 12.6 Interpolation inequality

If s=(1−θ)s0+θs1s = (1 - \theta)s_0 + \theta s_1 with 0≤θ≤10 \leq \theta \leq 1, then ∥u∥Hs≤∥u∥Hs01−θ∥u∥Hs1θ\|u\|_{H^s} \leq \|u\|_{H^{s_0}}^{1 - \theta}\|u\|_{H^{s_1}}^\theta.

Proof. (1+∣ξ∣2)s∣u^∣2=[(1+∣ξ∣2)s0∣u^∣2]1−θ[(1+∣ξ∣2)s1∣u^∣2]θ(1 + |\xi|^2)^s|\hat u|^2 = \big[(1 + |\xi|^2)^{s_0}|\hat u|^2\big]^{1 - \theta}\big[(1 + |\xi|^2)^{s_1}|\hat u|^2\big]^\theta; integrate and apply Hölder with exponents 11−θ\frac{1}{1 - \theta}, 1θ\frac1\theta.

So a function in L2L^2 and H2H^2 is in H1H^1 with ∥u∥H12≤∥u∥L2∥u∥H2\|u\|_{H^1}^2 \leq \|u\|_{L^2}\|u\|_{H^2}: one derivative is the geometric mean of none and two. Indeed HsH^s is the complex interpolation space [Hs0,Hs1]θ[H^{s_0}, H^{s_1}]_\theta in Calderón's sense: the three lines argument applied to the multiplier (1+∣ξ∣2)z/2(1 + |\xi|^2)^{z/2}.

Interpolation inequalities

The same idea, with LpL^p norms and derivatives, gives the inequalities used throughout geometric analysis to control intermediate quantities by extreme ones.

Proposition 12.7 Two one-dimensional inequalities

For u∈H1(R)u \in H^1(\mathbb{R}):

  1. ∥u∥∞2≤∥u∥2∥u′∥2\|u\|_\infty^2 \leq \|u\|_2\|u'\|_2;
  2. ∥u′∥22≤∥u∥2∥u′′∥2\|u'\|_2^2 \leq \|u\|_2\|u''\|_2 if also u∈H2u \in H^2.

Proof.

  1. For smooth compactly supported uu, u(x)2=∫−∞x2uu′=−∫x∞2uu′u(x)^2 = \int_{-\infty}^x2uu' = -\int_x^\infty2uu', so u(x)2≤∫∣uu′∣≤∥u∥2∥u′∥2u(x)^2 \leq \int|uu'| \leq \|u\|_2\|u'\|_2 (averaging the two formulas). Extend by density.
  2. ∫u′2=−∫uu′′\int u'^2 = -\int uu'' by parts, then Cauchy–Schwarz.

The second inequality is the special case s0=0s_0 = 0, s=1s = 1, s1=2s_1 = 2 of the previous proposition, in its homogeneous form; the first is a borderline case of Sobolev embedding (Hs⊆L∞H^s \subseteq L^\infty for s>12s > \frac12, 4A.6 The Fourier Transform) with the scaling-optimal split between L2L^2 and H1H^1.

Theorem 12.8 Gagliardo–Nirenberg interpolation inequality (statement)

Let 1≤q,r≤∞1 \leq q, r \leq \infty, j<mj < m integers, and jm≤a≤1\frac jm \leq a \leq 1 (with a<1a < 1 if m−j−nrm - j - \frac nr is a nonnegative integer). If

1p=jn+a(1r−mn)+1−aq,\frac1p = \frac jn + a\Big(\frac1r - \frac mn\Big) + \frac{1 - a}{q},

then ∥Dju∥Lp(Rn)≤C∥Dmu∥Lra∥u∥Lq1−a\|D^ju\|_{L^p(\mathbb{R}^n)} \leq C\|D^mu\|_{L^r}^a\|u\|_{L^q}^{1 - a} for uu in LqL^q with Dmu∈LrD^mu \in L^r.

The exponent relation is forced by scaling u(x)↦u(λx)u(x) \mapsto u(\lambda x), exactly as for Sobolev's exponent in 4A.11 Sobolev Embeddings and Critical Exponents (Exercise 12.13). Gagliardo and Nirenberg proved the general case independently in 1959. In geometric flows these inequalities are how one passes from L2L^2 bounds on curvature and its high derivatives to bounds on intermediate derivatives: Hamilton's derivative estimates for the Ricci flow on compact manifolds use interpolation inequalities of exactly this kind (11A.5 Shi's Derivative Estimates), as do the regularity theories for mean curvature flow and harmonic map flow.

Traces

On a half-space R+n={xn>0}\mathbb{R}^n_+ = \{x_n > 0\}, a function in H1H^1 has no well-defined values on a set of measure zero in general, but its restriction to the hyperplane {xn=0}\{x_n = 0\} makes sense in a weaker space. The Fourier transform identifies it precisely.

Theorem 12.9 The trace theorem

Let s>12s > \frac12. The restriction γu(x′)=u(x′,0)\gamma u(x') = u(x', 0), defined for u∈S(Rn)u \in \mathcal S(\mathbb{R}^n), extends to a bounded operator γ:Hs(Rn)→Hs−1/2(Rn−1)\gamma : H^s(\mathbb{R}^n) \to H^{s - 1/2}(\mathbb{R}^{n - 1}), which is onto.

Proof. Boundedness. For Schwartz uu, Fourier inversion in the last variable gives γu^(ξ′)=∫Ru^(ξ′,ξn) dξn\widehat{\gamma u}(\xi') = \int_{\mathbb{R}}\hat u(\xi', \xi_n)\,d\xi_n. By Cauchy–Schwarz,

∣γu^(ξ′)∣2≤∫(1+∣ξ∣2)s∣u^(ξ′,ξn)∣2dξn⋅∫dξn(1+∣ξ′∣2+ξn2)s.|\widehat{\gamma u}(\xi')|^2 \leq \int(1 + |\xi|^2)^s|\hat u(\xi', \xi_n)|^2d\xi_n\cdot\int\frac{d\xi_n}{(1 + |\xi'|^2 + \xi_n^2)^s}.

With ξn=(1+∣ξ′∣2)1/2t\xi_n = (1 + |\xi'|^2)^{1/2}t, the second integral is (1+∣ξ′∣2)12−s∫dt(1+t2)s=cs(1+∣ξ′∣2)12−s(1 + |\xi'|^2)^{\frac12 - s}\int\frac{dt}{(1 + t^2)^s} = c_s(1 + |\xi'|^2)^{\frac12 - s}, finite exactly because s>12s > \frac12. Multiplying by (1+∣ξ′∣2)s−12(1 + |\xi'|^2)^{s - \frac12} and integrating over ξ′\xi' gives ∥γu∥Hs−1/22≤cs∥u∥Hs2\|\gamma u\|_{H^{s - 1/2}}^2 \leq c_s\|u\|_{H^s}^2. Extend by density.

Onto. Given g∈Hs−1/2(Rn−1)g \in H^{s - 1/2}(\mathbb{R}^{n - 1}), define uu by u^(ξ′,ξn)=g^(ξ′)ψ(ξn/⟨ξ′⟩)⟨ξ′⟩\hat u(\xi', \xi_n) = \hat g(\xi')\frac{\psi(\xi_n/\langle\xi'\rangle)}{\langle\xi'\rangle} with ⟨ξ′⟩=(1+∣ξ′∣2)1/2\langle\xi'\rangle = (1 + |\xi'|^2)^{1/2} and ψ\psi smooth, compactly supported, ∫ψ=1\int\psi = 1. Then ∫u^ dξn=g^\int\hat u\,d\xi_n = \hat g, and ∥u∥Hs2≤C∫∣g^∣2⟨ξ′⟩2s−2⋅⟨ξ′⟩ dξ′=C∥g∥Hs−1/22\|u\|_{H^s}^2 \leq C\int|\hat g|^2\langle\xi'\rangle^{2s - 2}\cdot\langle\xi'\rangle\,d\xi' = C\|g\|_{H^{s - 1/2}}^2 (Exercise 12.14).

So the trace space of H1H^1 is H1/2H^{1/2}: a function on the boundary is the boundary value of an H1H^1 function exactly when it has half a derivative in L2L^2. On a bounded smooth domain the same holds, with H1/2(∂Ω)H^{1/2}(\partial\Omega) defined by the Gagliardo seminorm on ∂Ω\partial\Omega (or by local charts), and H01(Ω)H^1_0(\Omega) is exactly the kernel of the trace (4A.10 Sobolev Spaces). This completes the weak theory of the Dirichlet problem Δu=0\Delta u = 0, u=gu = g on ∂Ω\partial\Omega: it has a unique solution in H1(Ω)H^1(\Omega) for every g∈H1/2(∂Ω)g \in H^{1/2}(\partial\Omega) (extend gg to G∈H1G \in H^1, then solve for u−G∈H01u - G \in H^1_0 by Lax–Milgram, 4A.5 Hilbert Spaces and Lax–Milgram). For W1,pW^{1,p} the trace space is W1−1/p,p(∂Ω)W^{1 - 1/p,p}(\partial\Omega) (Gagliardo, 1957), and for p=1p = 1 it is all of L1(∂Ω)L^1(\partial\Omega).

The fractional Laplacian

For 0<s<10 < s < 1 define (−Δ)s(-\Delta)^s by its symbol: (−Δ)su^(ξ)=(2π∣ξ∣)2su^(ξ)\widehat{(-\Delta)^su}(\xi) = (2\pi|\xi|)^{2s}\hat u(\xi). Its quadratic form is the homogeneous HsH^s seminorm, which by Proposition 12.5 is the Gagliardo seminorm, and polarising gives the singular integral formula

(−Δ)su(x)=cn,s P.V. ⁣∫Rnu(x)−u(y)∣x−y∣n+2s dy(-\Delta)^su(x) = c_{n,s}\,\mathrm{P.V.}\!\int_{\mathbb{R}^n}\frac{u(x) - u(y)}{|x - y|^{n + 2s}}\,dy

(Exercise 12.15). The value at xx depends on uu everywhere: (−Δ)s(-\Delta)^s is nonlocal. Probabilistically, −(−Δ)s-(-\Delta)^s is the generator of the symmetric 2s2s-stable Lévy process, whose jumps of length rr have intensity proportional to r−n−2sr^{-n - 2s}: exactly the kernel in the integral. Its heat kernel decays like ∣x∣−n−2s|x|^{-n - 2s}, polynomially, rather than like a Gaussian, which is the heavy tail of a Lévy flight.

Luis Caffarelli and Luis Silvestre showed in 2007 that (−Δ)s(-\Delta)^s is a Dirichlet-to-Neumann map: if U(x,y)U(x, y) on Rn×(0,∞)\mathbb{R}^n\times(0, \infty) solves div⁡(y1−2s∇U)=0\operatorname{div}(y^{1 - 2s}\nabla U) = 0 with U(x,0)=u(x)U(x, 0) = u(x), then (−Δ)su=−clim⁡y→0y1−2s∂yU(-\Delta)^su = -c\lim_{y\to0}y^{1 - 2s}\partial_yU. This extension turns nonlocal problems into local, degenerate elliptic ones in one more dimension, and opened the regularity theory of nonlocal equations. For s=12s = \frac12 it is the classical fact that (−Δ)1/2(-\Delta)^{1/2} is the Dirichlet-to-Neumann map of the half-space, the operator behind the trace theorem: ∥g∥H˙1/22\|g\|_{\dot H^{1/2}}^2 is, up to a constant, the minimal Dirichlet energy of an extension of gg (Exercise 12.16).

In the world In use Fractional smoothness in image and signal processing

Image and signal processing measures the regularity of signals by fractional smoothness. Wavelet coefficients decay at rates that characterise Sobolev and Besov spaces of fractional order, and compression schemes, including the wavelet-based JPEG 2000 standard, are most efficient on images whose fractional smoothness is high away from edges. Denoising by thresholding wavelet coefficients (Donoho and Johnstone, 1994) is provably near-optimal over Besov classes. The point in each case is that natural images are not in H1H^1 (edges have infinite Dirichlet energy) but are in HsH^s for s<12s < \frac12, and in Besov spaces adapted to edges.

Where this goes Where fractional smoothness returns

Traces in H1/2H^{1/2} complete the Dirichlet problem used in 6B.7 Schauder Theory and the Continuity Method and 6C.6 Quasilinear Elliptic Equations and Minimal Surfaces; LpL^p–LqL^q estimates for the heat semigroup from Young's inequality drive 6A.8 Nonlinear Parabolic Equations; Marcinkiewicz interpolation and singular integrals give the W2,pW^{2,p} theory of 6B.8 Lᵖ Theory and Strong Solutions. 4A.13 Unbounded Operators and the Spectral Theorem turns from function spaces to operators: the Laplacian, (−Δ)s(-\Delta)^s and −4Δ+R-4\Delta + R are unbounded operators, and their functions etΔe^{t\Delta}, (−Δ)s(-\Delta)^s, (−Δ)−1(-\Delta)^{-1} are defined by the spectral theorem.

History

Marcel Riesz's interpolation theorem dates from 1926 and Thorin's complex proof from 1939, Marcinkiewicz's real interpolation theorem from the same year (published posthumously, Zygmund 1956). Hausdorff (1923) and Young (1912) proved the Fourier inequalities named after them for Fourier series. Fractional Sobolev spaces were developed by Aronszajn, Gagliardo and Slobodeckij in the 1950s; Gagliardo's trace theorem appeared in 1957, the Gagliardo–Nirenberg inequalities in 1959, and the abstract theories of complex and real interpolation of Banach spaces (Calderón; Lions and Peetre) in the early 1960s. Caffarelli and Silvestre's extension paper is from 2007.

Recall Where we stand

The three lines lemma (log-convexity of the maximum of a bounded holomorphic function across a strip) gives Riesz–Thorin: an operator bounded Lp0→Lq0L^{p_0} \to L^{q_0} and Lp1→Lq1L^{p_1} \to L^{q_1} is bounded at every convex combination of the reciprocal exponents, with interpolated norm. Corollaries: Hausdorff–Young ∥f^∥p′≤∥f∥p\|\hat f\|_{p'} \leq \|f\|_p for p≤2p \leq 2, and Young's inequality ∥f∗g∥q≤∥f∥p∥g∥r\|f*g\|_q \leq \|f\|_p\|g\|_r, which gives LpL^p–LqL^q smoothing for the heat equation. HsH^s is defined by (1+∣ξ∣2)s/2u^∈L2(1 + |\xi|^2)^{s/2}\hat u \in L^2; for 0<s<10 < s < 1 it equals the space defined by the Gagliardo seminorm, and its norms interpolate by Hölder. Interpolation inequalities (one derivative is the geometric mean of none and two; Gagliardo–Nirenberg in general) control intermediate derivatives. Traces of HsH^s functions lie in Hs−1/2H^{s - 1/2}, so boundary values of H1H^1 functions are exactly H1/2H^{1/2}. The fractional Laplacian has symbol ∣2πξ∣2s|2\pi\xi|^{2s}, is a nonlocal singular integral, generates Lévy flights, and is a Dirichlet-to-Neumann map by the Caffarelli–Silvestre extension.

Exercises

Exercise 12.10 Riesz–Thorin at the ends

(a) Check that ∣fit∣=∣f∣pθ/p0|f_{it}| = |f|^{p_\theta/p_0} and hence ∥fit∥p0=1\|f_{it}\|_{p_0} = 1 when ∥f∥pθ=1\|f\|_{p_\theta} = 1. (b) If p0=∞p_0 = \infty, how should fzf_z be modified? (c) Prove the scalar analogue ∥f∥pθ≤∥f∥p01−θ∥f∥p1θ\|f\|_{p_\theta} \leq \|f\|_{p_0}^{1 - \theta}\|f\|_{p_1}^\theta (log-convexity of LpL^p norms) directly with Hölder's inequality.

Solution

(a) ∣fit∣=∣f∣pθRe⁡(1−itp0+itp1)=∣f∣pθ/p0|f_{it}| = |f|^{p_\theta\operatorname{Re}(\frac{1 - it}{p_0} + \frac{it}{p_1})} = |f|^{p_\theta/p_0}, so ∫∣fit∣p0=∫∣f∣pθ=1\int|f_{it}|^{p_0} = \int|f|^{p_\theta} = 1. (b) With 1p0=0\frac{1}{p_0} = 0 the formula gives ∣fit∣=∣f∣0=1|f_{it}| = |f|^{0} = 1 on the support of ff: fit=eiαf_{it} = e^{i\alpha} there, of sup norm 11. So the same formula works with the convention 1∞=0\frac1\infty = 0. (c) Write ∣f∣pθ=∣f∣(1−θ)pθ∣f∣θpθ|f|^{p_\theta} = |f|^{(1 - \theta)p_\theta}|f|^{\theta p_\theta} with exponents p0(1−θ)pθ\frac{p_0}{(1 - \theta)p_\theta} and p1θpθ\frac{p_1}{\theta p_\theta}, whose reciprocals sum to (1−θ)pθp0+θpθp1=1\frac{(1 - \theta)p_\theta}{p_0} + \frac{\theta p_\theta}{p_1} = 1; Hölder gives ∫∣f∣pθ≤∥f∥p0(1−θ)pθ∥f∥p1θpθ\int|f|^{p_\theta} \leq \|f\|_{p_0}^{(1 - \theta)p_\theta}\|f\|_{p_1}^{\theta p_\theta}. This is the scalar shadow of Riesz–Thorin.

Exercise 12.11 No Hausdorff–Young for p>2p > 2

(a) On R\mathbb{R}, use complex Gaussians ga(x)=e−πax2g_a(x) = e^{-\pi ax^2} with complex aa, Re⁡a>0\operatorname{Re}a > 0, whose transform is a−1/2e−πξ2/aa^{-1/2}e^{-\pi\xi^2/a}. Taking a=ε+ia = \varepsilon + i, compute ∥ga∥p\|g_a\|_p and ∥g^a∥p′\|\hat g_a\|_{p'} as ε→0\varepsilon \to 0 and show the ratio ∥g^a∥p′/∥ga∥p\|\hat g_a\|_{p'}/\|g_a\|_p is unbounded when p>2p > 2. (b) Why does this not contradict the case p≤2p \leq 2?

Solution

(a) ∣ga∣=e−πεx2|g_a| = e^{-\pi\varepsilon x^2}, so ∥ga∥p=(pε)−1/(2p)\|g_a\|_p = (p\varepsilon)^{-1/(2p)}. 1a=ε−iε2+1\frac1a = \frac{\varepsilon - i}{\varepsilon^2 + 1}, so ∣g^a∣=∣a∣−1/2e−πεξ2/(1+ε2)|\hat g_a| = |a|^{-1/2}e^{-\pi\varepsilon\xi^2/(1 + \varepsilon^2)}, and ∥g^a∥p′≈(p′ε)−1/(2p′)\|\hat g_a\|_{p'} \approx \big(p'\varepsilon\big)^{-1/(2p')}. The ratio behaves like ε12p−12p′\varepsilon^{\frac{1}{2p} - \frac{1}{2p'}}, which tends to ∞\infty exactly when 1p<1p′\frac1p < \frac1{p'}, that is p>2p > 2. (b) For p<2p < 2 the same ratio tends to 00, consistent with the bound; the chirps eiπx2e^{i\pi x^2} spread a function out in frequency less than in space when p>2p > 2.

Exercise 12.12 Heat semigroup on LpL^p

Using ∥Φt∥r=crt−n2(1−1r)\|\Phi_t\|_r = c_rt^{-\frac n2(1 - \frac1r)} for the heat kernel Φt(x)=(4πt)−n/2e−∣x∣2/4t\Phi_t(x) = (4\pi t)^{-n/2}e^{-|x|^2/4t}, derive ∥etΔf∥q≤Ct−n2(1p−1q)∥f∥p\|e^{t\Delta}f\|_q \leq Ct^{-\frac n2(\frac1p - \frac1q)}\|f\|_p for 1≤p≤q≤∞1 \leq p \leq q \leq \infty. Check the exponent by scaling.

Solution

Young with 1+1q=1p+1r1 + \frac1q = \frac1p + \frac1r: ∥Φt∗f∥q≤∥Φt∥r∥f∥p\|\Phi_t*f\|_q \leq \|\Phi_t\|_r\|f\|_p, and 1−1r=1p−1q1 - \frac1r = \frac1p - \frac1q. Scaling: if fλ(x)=f(λx)f_\lambda(x) = f(\lambda x) then etΔfλ=(eλ2tΔf)(λ⋅)e^{t\Delta}f_\lambda = (e^{\lambda^2t\Delta}f)(\lambda\cdot), and ∥g(λ⋅)∥q=λ−n/q∥g∥q\|g(\lambda\cdot)\|_q = \lambda^{-n/q}\|g\|_q; matching powers of λ\lambda on both sides forces the exponent −n2(1p−1q)-\frac n2(\frac1p - \frac1q).

Exercise 12.13 Scaling and Gagliardo–Nirenberg

Replace uu by uλ(x)=u(λx)u_\lambda(x) = u(\lambda x) in ∥Dju∥p≤C∥Dmu∥ra∥u∥q1−a\|D^ju\|_p \leq C\|D^mu\|_r^a\|u\|_q^{1 - a} and show the inequality can hold for all λ\lambda only if 1p=jn+a(1r−mn)+1−aq\frac1p = \frac jn + a(\frac1r - \frac mn) + \frac{1 - a}{q}. Specialise to n=1n = 1, j=0j = 0, m=1m = 1, p=∞p = \infty, q=r=2q = r = 2 and recover a=12a = \frac12, the first inequality of Proposition 12.7.

Solution

∥Dkuλ∥s=λk−n/s∥Dku∥s\|D^ku_\lambda\|_s = \lambda^{k - n/s}\|D^ku\|_s. Equating exponents: j−np=a(m−nr)+(1−a)(−nq)j - \frac np = a(m - \frac nr) + (1 - a)(-\frac nq); divide by −n-n. For the special case: 0=a(12−1)+1−a20 = a(\frac12 - 1) + \frac{1 - a}{2}, so a=12a = \frac12.

Exercise 12.14 Extending boundary data

In the proof of the trace theorem, verify ∫u^ dξn=g^(ξ′)\int\hat u\,d\xi_n = \hat g(\xi') and ∥u∥Hs2≤C∥g∥Hs−1/22\|u\|_{H^s}^2 \leq C\|g\|_{H^{s - 1/2}}^2. (Substitute ξn=⟨ξ′⟩t\xi_n = \langle\xi'\rangle t and note 1+∣ξ∣2=⟨ξ′⟩2(1+t2)1 + |\xi|^2 = \langle\xi'\rangle^2(1 + t^2).)

Solution

∫g^⟨ξ′⟩−1ψ(ξn/⟨ξ′⟩) dξn=g^∫ψ=g^\int\hat g\langle\xi'\rangle^{-1}\psi(\xi_n/\langle\xi'\rangle)\,d\xi_n = \hat g\int\psi = \hat g. And ∫(1+∣ξ∣2)s∣u^∣2dξn=∣g^∣2⟨ξ′⟩−2∫⟨ξ′⟩2s(1+t2)sψ(t)2⟨ξ′⟩ dt=Cψ∣g^∣2⟨ξ′⟩2s−1\int(1 + |\xi|^2)^s|\hat u|^2d\xi_n = |\hat g|^2\langle\xi'\rangle^{-2}\int\langle\xi'\rangle^{2s}(1 + t^2)^s\psi(t)^2\langle\xi'\rangle\,dt = C_\psi|\hat g|^2\langle\xi'\rangle^{2s - 1}, whose integral over ξ′\xi' is Cψ∥g∥Hs−1/22C_\psi\|g\|_{H^{s - 1/2}}^2.

Exercise 12.15 The singular integral formula

For u∈S(Rn)u \in \mathcal S(\mathbb{R}^n) and 0<s<10 < s < 1, show that 12∫2u(x)−u(x+h)−u(x−h)∣h∣n+2sdh\frac12\int\frac{2u(x) - u(x + h) - u(x - h)}{|h|^{n + 2s}}dh has Fourier transform 12I(ξ)u^(ξ)\frac12 I(\xi)\hat u(\xi), with II as in Proposition 12.5, and conclude that it is a constant multiple of (−Δ)su(-\Delta)^su. Why does the symmetrised second difference make the integral converge without a principal value?

Solution

The transform of 2u(x)−u(x+h)−u(x−h)2u(x) - u(x + h) - u(x - h) is (2−e2πih⋅ξ−e−2πih⋅ξ)u^=∣e2πih⋅ξ−1∣2u^(2 - e^{2\pi ih\cdot\xi} - e^{-2\pi ih\cdot\xi})\hat u = |e^{2\pi ih\cdot\xi} - 1|^2\hat u. Integrating against ∣h∣−n−2s|h|^{-n - 2s} gives I(ξ)u^=I(e1)∣ξ∣2su^I(\xi)\hat u = I(e_1)|\xi|^{2s}\hat u. Near h=0h = 0 the second difference is O(∣h∣2)O(|h|^2) by Taylor (the first-order terms cancel), so the integrand is O(∣h∣2−n−2s)O(|h|^{2 - n - 2s}), integrable since s<1s < 1.

Exercise 12.16 Rehearsal: H1/2H^{1/2} as Dirichlet energy

For g∈S(Rn−1)g \in \mathcal S(\mathbb{R}^{n - 1}), the harmonic extension to R+n\mathbb{R}^n_+ is U^(ξ′,y)=g^(ξ′)e−2π∣ξ′∣y\hat U(\xi', y) = \hat g(\xi')e^{-2\pi|\xi'|y} (Fourier transform in x′x'). Show that ∫R+n∣∇U∣2=2π∫∣ξ′∣∣g^∣2dξ′\int_{\mathbb{R}^n_+}|\nabla U|^2 = 2\pi\int|\xi'||\hat g|^2d\xi', a multiple of the homogeneous H˙1/2\dot H^{1/2} seminorm. Why is the harmonic extension the energy minimiser among extensions?

Solution

∣∇x′U^∣2+∣∂yU^∣2=(4π2∣ξ′∣2+4π2∣ξ′∣2)∣g^∣2e−4π∣ξ′∣y|\widehat{\nabla_{x'}U}|^2 + |\widehat{\partial_yU}|^2 = (4\pi^2|\xi'|^2 + 4\pi^2|\xi'|^2)|\hat g|^2e^{-4\pi|\xi'|y}; integrating in yy gives 8π2∣ξ′∣2∣g^∣2⋅14π∣ξ′∣=2π∣ξ′∣∣g^∣28\pi^2|\xi'|^2|\hat g|^2\cdot\frac{1}{4\pi|\xi'|} = 2\pi|\xi'||\hat g|^2; integrate over ξ′\xi' by Plancherel. Harmonic functions minimise Dirichlet energy with given boundary values (Dirichlet's principle, 6A.2 Harmonic Functions): for any other extension V=U+WV = U + W with W=0W = 0 on the boundary, ∫∣∇V∣2=∫∣∇U∣2+∫∣∇W∣2\int|\nabla V|^2 = \int|\nabla U|^2 + \int|\nabla W|^2, since the cross term 2∫∇U⋅∇W=−2∫WΔU=02\int\nabla U\cdot\nabla W = -2\int W\Delta U = 0.

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