PARTIAL DIFFERENTIAL EQUATIONS

IMPA - AUG-NOV 2011

LECTURE 2

EMANUEL CARNEIRO

1. Interpolation techniques

1.1. The Riesz-Thorin interpolation. Let us first recall some basic facts aboutfunctions and operators in Lp-spaces.

Proposition 1. Let 1 ≤ p < q ≤ ∞. If f ∈ Lp(Rn) ∩ Lq(Rn) then f ∈ Lr(Rn), forany p ≤ r ≤ q. In fact, if we write

1r

=θ

p+

1− θq

,

with 0 ≤ θ ≤ 1, we have the inequality

‖f‖Lr(Rn) ≤ ‖f‖θLp(Rn) ‖f‖1−θLq(Rn).

Proof. Using Holder’s inequality we have

‖f‖Lr(Rn) =(∫

Rn

|f(x)|r dx)1/r

=(∫

Rn

|f(x)|rθ |f(x)|r(1−θ)dx)1/r

≤(∫

Rn

|f(x)|p dx)θ/p (∫

Rn

|f(x)|q dx)(1−θ)/p

= ‖f‖θLp(Rn) ‖f‖1−θLq(Rn).

�

Next we prove a version for operators of the last proposition, a result called theRiesz-Thorin interpolation theorem. First we need a lemma from complex analysis.

Lemma 2 (Hadamard’s three lines lemma). Let Φ be a bounded and continuous func-tion on the strip 0 ≤ <z ≤ 1 that is holomorphic on the interior of the strip. If|Φ(z)| ≤ M0 for <(z) = 0 and |Φ(z)| ≤ M1 for <(z) = 1 then |Φ(z)| ≤ M1−t

0 M t1 for

<(z) = t with 0 < t < 1.

Proof. For ε > 0 consider the function

Φε(z) =e−εz(1−z)Φ(z)M1−z

0 Mz1

Observe that Φε satisfies the hypotheses of the lemma with M0 = M1 = 1 andthat |Φε| → 0 as =(z) → ∞. Therefore |Φε(z)| ≤ 1 on the boundary of the rectangle

Date: August 31, 2011.2000 Mathematics Subject Classification. XX-XXX.Key words and phrases. XXX-XXX.

1

2 EMANUEL CARNEIRO

0 ≤ <(z) ≤ 1 and −A ≤ =(z) ≤ A (for A large) and the maximum modulus principleimplies that |Φε(z)| ≤ 1 on the strip 0 ≤ <(z) ≤ 1. Letting ε→ 0 we obtain

|Φ(z)|M t−10 M−t1 = lim

ε→0|Φε(z)| ≤ 1,

for <(z) = t, and this concludes the proof. �

Theorem 3 (Riesz-Thorin interpolation). Let 1 ≤ p0, q0, p1, q1 ≤ ∞. Suppose thatT is a linear map from Lp0(Rn) + Lp1(Rn) into Lq0(Rn) + Lq1(Rn) that satisfies thefollowing estimates

‖Tf‖Lq0 (Rn) ≤M0‖f‖Lp0 (Rn)

and‖Tf‖Lq1 (Rn) ≤M1‖f‖Lp1 (Rn).

Let (pt, qt) be a point such that

1pt

=1− tp0

+t

p1and

1qt

=1− tq0

+t

q1,

for some 0 < t < 1 (i.e. the point (1/pt, 1/qt) belongs to the segment connecting(1/p0, 1/q0) and (1/p1, 1/q1)). Therefore we have

‖Tf‖Lt (Rn) ≤M1−t0 M t

1 ‖f‖Lpt (Rn). (1.1)

Proof. The case p0 = p1 is given by Proposition 1. Now suppose that p0 6= p1 and thusthat 1 < pt <∞. We shall prove first that inequality (1.1) holds for f in the space Σof simple functions that vanish outside a set of finite measure (note that Σ is dense inLp(Rn), for 1 ≤ p <∞). For this observe that by duality we have

‖Tf‖Lqt (Rn) = sup{∣∣∣∣∫

Rn

Tf(x) g(x) dx∣∣∣∣ , g ∈ Σ, ‖g‖

Lq′t = 1}

We may assume that f 6= 0 and rescale to get ‖f‖Lpt (Rn) = 1. Write

f =m∑1

cjχEjand g =

n∑1

dkχFk,

where the sets Ej are disjoint and the sets Fk are disjoint as well, and the cj ’s and dk’sare non-zero. Write cj and dk in the polar form

cj = |cj |eiθj and dk = |dk|eiψk .

Let

α(z) =1− zp0

+z

p1and β(z) =

1− zq0

+z

q1,

and thus α(t) = p−1t and β(t) = q−1

t , for 0 < t < 1. Fix t ∈ (0, 1) and since we haveassumed pt <∞ we have α(t) > 0. We may then define

fz =m∑1

|cj |α(z)/α(t)eiθjχEj .

If β(t) < 1 we define

gz =n∑1

|dk|(1−β(z))/(1−β(t))eiψkχFk,

PDE - LECTURE 2 3

while if β(t) = 1 we set gz = g for all z. We henceforth assume β(t) < 1, since theother is a simple modification of the argument. Finally we set

Φ(z) =∫

Rn

Tfz(x) gz(x) dx.

ThereforeΦ(z) =

∑j,k

Ajk|cj |α(z)/α(t)|dk|(1−β(z))/(1−β(t)),

whereAjk = ei(θj+ψk)

∫Rn

TχEj(x)χFk

(x) dx.

and thus we see that Φ is and entire function that is bounded on the strip 0 ≤ <(z) ≤ 1.Since

∫(Tf)gdx = Φ(t), by the three lines lemma it suffices to show that |Φ(z)| ≤M0

when <(z) = 0, and |Φ(z)| ≤M1 when <(z) = 1. Now observe that

α(is) =1p0

+ is

(1p1− 1p0

)and 1− β(is) =

(1− 1

q0

)− is

(1q1− 1q0

),

for s ∈ R and therefore

|fis| = |f |<(α(is)/α(t)) = |f |pt/p0 and |gis| = |g|<[(1−β(is))/(1−α(t))] = |g|q′t/q

′0 .

By Holder’s inequality we have

|Φ(is)| ≤ ‖Tfis‖Lq0 ‖gis‖Lq′0≤M0‖fis‖Lp0 ‖gis‖Lq′0

= M0‖f‖pt/p0Lpt ‖g‖

q′t/q′0

Lq′t= M0.

A similar calculation shows that |Φ(1 + is)| ≤ M1, and we have proved that (1.1)holds for f ∈ Σ. For the general case, given f ∈ Lpt(Rn), we can choose a sequence{fn} in Σ such that |fn| ≤ |f | and fn → f pointwise. Also, let E = {x : |f(x)| > 1,g = fχE , gn = fnχE , h = f − g, hn = fn− gn. Suppose without loss of generality thatp0 < p1, then g ∈ Lp0 and h ∈ Lp1 , and by dominated convergence ‖fn − f‖Lpt → 0,‖gn−g‖Lp0 → 0 and ‖hn−h‖Lp1 → 0. Hence ‖Tgn−Tg‖Lq0 → 0 and ‖Thn−Th‖Lq1 →0, and by passing to a suitable subsequence we may assume that Tgn → Tg a.e andThn → Th a.e.. But then Tfn → Tf a.e. and by Fatou’s lemma

‖Tf‖Lqt ≤ lim inf ‖Tfn‖Lqt ≤ lim inf M1−t0 M t

1‖fn‖Lpt = M1−t0 M t

1‖f‖Lpt ,

and this concludes the proof. �

1.2. Two applications. We list two classical applications of this result. First definethe convolution of two functions f and g by

f ∗ g(x) =∫

Rn

f(y) (g(x− y) dy =∫

Rn

f(x− y) (g(y) dy.

Theorem 4 (Young’s inequality for convolutions). Let 1 ≤ p, q, r ≤ ∞. Then

‖f ∗ g‖Lr(Rn) ≤ ‖f‖Lp(Rn))‖g‖Lq(Rn).

Proof. One can see the proof of this theorem in three ways. First, one could see that theproposed inequality is true for (p, q, r) = (1, 1, 1), (1,∞,∞) and (∞, 1,∞) and appealto a multilinear (in this case bilinear) version on the Riesz-Thorin intepolation to fillin the convex hull. The proof of a multilinear version of the Riesz-Thorin interpolationis basically the same as we did above.

Second, if one wants to use the theorem as we proved above, we should proceed thisway. If is easy to se that we can have (p, q, r) = (1, 1, 1). From Holder’s inequality wealso have it for (p, q, r) = (p, p′,∞). In particular fixing f ∈ L1 we can see that this

4 EMANUEL CARNEIRO

is a linear operator in g and from the triples (1, 1, 1) and (1,∞,∞) we can produce(1, q, q). Now fix g ∈ Lq and look this linear operator in f . From the points (1, q, q)and (q′, q,∞) we reach the desirable (p, q, r) satisfying 1 = 1/r = 1/p+ 1/q.

Third, one could prove this theorem directly from a clever use of Holder’s inequality.I will leave this as an exercise. �

Now define the Fourier transform of a function f ∈ L1(Rn) by

f(ξ) =∫

Rn

e−2πix·ξ f(x) dx.

It is clear from the definition that

‖f‖L∞(Rn) ≤ ‖f‖L1(Rn).

We shall prove later in this course that the Fourier transform can be extended in naturalway to L2(Rn) and satisfies Plancherel’s theorem

‖f‖L2(Rn) = ‖f‖L2(Rn).

From these two points, using Riesz interpolation we have

Theorem 5 (Hausdorff-Young’s inequality for the Fourier transform). Let 1 ≤ p ≤ 2.Then

‖f‖Lp′ (Rn) ≤ ‖f‖Lp(Rn).

References

[1] L. C. Evans, Partial Differential Equations, Graduate Studies in Mathematics, AMS, Volume 19,

1998.

[2] G. B. Folland, Real Analysis, Modern Techniques and Their Applications, John Wiley and Sons,1999.

[3] R. Iorio and V. M. Iorio, Fourier Analysis and Partial Differential Equations, Cambridge studies

in advanced mathematics, Cambridge University Press, 2001.

IMPA - Estrada Dona Castorina, 110, Rio de Janeiro, RJ, Brazil 22460-320

E-mail address: carneiro@impa.br