1.2 Convolution and Approximate Identities
1.2.4 Approximate Identities
Fig. 1.2 The Fej´er kernel F5 plotted on the interval [−12,12].
1 π Z
|x|≥δ
1 ε
1
(x/ε)2+1dx=1−2
πarctan(δ/ε)→0 asε→0, for allδ>0. The functionPεis called thePoisson kernel.
Example 1.2.18.On the circle groupT1let FN(t) =
N j=−N
∑
1− |j|
N+1
e2πi jt= 1 N+1
sin(π(N+1)t) sin(πt)
2
. (1.2.27) To check the previous equality we use that
sin2(x) = (2−e2ix−e−2ix)/4,
and we carry out the calculation.FN is called theFej´er kernel. To see that the se- quence{FN}N is an approximate identity, we check conditions (i), (ii), and (iii) in Definition 1.2.15. Property (iii) follows from the expression givingFN in terms of sines, while property (i) follows from the expression givingFNin terms of exponen- tials. Property (ii) is identical to property (i), sinceFNis nonnegative.
Next comes the basic theorem concerning approximate identities.
Theorem 1.2.19.Let kε be an approximate identity on a locally compact group G with left Haar measureλ.
(1) If f ∈Lp(G)for1≤p<∞, then
kε∗f−f
Lp(G)→0asε→0.
(2) When p=∞, the following is valid: If f is continuous in a neighborhood of a a compact subset K of G, then
kε∗f−f
L∞(K)→0asε→0.
Proof. We start with the case 1≤p<∞. We recall that continuous functions with compact support are dense in Lp of locally compact Hausdorff spaces equipped with measures arising from nonnegative linear functionals (see Hewitt and Ross [125], Theorem 12.10). For a continuous function with compact supportgwe have
|g(h−1x)−g(x)|p≤(2 g
L∞)p forhin a relatively compact neighborhood of the origine, and by the Lebesgue dominated convergence theorem we obtain
-0.4 -0.2 0.2 0.4
1 2 3 4 5 6
Z
G
|g(h−1x)−g(x)|pdλ(x)→0 (1.2.28) ash→e. Now approximate a givenf inLp(G)by a continuous function with com- pact supportgto deduce that
Z
G
|f(h−1x)−f(x)|pdλ(x)→0 as h→e. (1.2.29) Because of (1.2.29), given aδ >0 there exists a neighborhoodV ofesuch that
h∈V =⇒ Z
G
|f(h−1x)−f(x)|pdλ(x)<
δ 2c
p
, (1.2.30)
wherecis the constant that appears in Definition 1.2.15 (i). Sincekε has integral one for allε>0, we have
(kε∗f)(x)−f(x) = (kε∗f)(x)−f(x) Z
G
kε(y)dλ(y)
= Z
G
(f(y−1x)−f(x))kε(y)dλ(y)
= Z
V
(f(y−1x)−f(x))kε(y)dλ(y) +
Z
Vc
(f(y−1x)−f(x))kε(y)dλ(y).
(1.2.31)
Now takeLpnorms inxin (1.2.31). In view of (1.2.30),
Z
V
(f(y−1x)−f(x))kε(y)dλ(y)
Lp(G,dλ(x))
≤ Z
V
f(y−1x)−f(x)
Lp(G,dλ(x))|kε(y)|dλ(y)
≤ Z
V
δ
2c|kε(y)|dλ(y)<δ 2,
(1.2.32)
while Z
Vc
(f(y−1x)−f(x))kε(y)dλ(y)
Lp(G,dλ(x))
≤ Z
Vc
2 f
Lp(G)|kε(y)|dλ(y)<δ 2,
(1.2.33)
provided we have that Z
Vc
|kε(x)|dλ(x)< δ 4
f Lp
. (1.2.34)
Chooseε0>0 such that (1.2.34) is valid forε<ε0by property (iii). Now (1.2.32) and (1.2.33) imply the required conclusion.
The case p=∞follows similarly. Since f is uniformly continuous onK, given δ >0 find a neighborhoodV ofe∈Gsuch that
h∈V =⇒ |f(h−1x)−f(x)|< δ
2c for allx∈K, (1.2.35)
wherecis as in Definition 1.2.15(i), and then find anε0>0 such that for 0<ε<ε0
we have
Z
Vc
|kε(y)|dλ(y)< δ 4
f L∞
. (1.2.36)
Using (1.2.35) and (1.2.36), we easily conclude that sup
x∈K
|(kε∗f)(x)−f(x)|
≤ Z
V
|kε(y)|sup
x∈K
|f(y−1x)−f(x)|dλ(y) + Z
Vc
|kε(y)|sup
x∈K
|f(y−1x)−f(x)|dλ(y)
≤ δ 2+δ
2 =δ,
which shows thatkε∗f converges uniformly to f onKasε→0.
Remark 1.2.20.Observe that if Haar measure satisfies (1.2.12), then the conclusion of Theorem 1.2.19 also holds for f∗kε.
A simple modification in the proof of Theorem 1.2.19 yields the following vari- ant.
Theorem 1.2.21.Let kεbe a family of functions on a locally compact group G that satisfies properties (i) and (iii) of Definition 1.2.15 and also
Z
G
kε(x)dλ(x) =a∈C, for allε>0. Let f ∈Lp(G)for some1≤p≤∞.
(a) If1≤p<∞, then
kε∗f−a f
Lp(G)→0asε→0. (b) If p=∞and f is continuous on a compact K⊆G, then
kε∗f−a f
L∞(K)→0 asε→0.
Remark 1.2.22.With the notation of Theorem 1.2.21, if f is continuous and tends to zero at infinity, then
kε∗f−a f
L∞(G) →0. To see this, simply observe that outside a compact subset ofG, bothkε∗f,a f have smallL∞norm, while inside a compact subset ofG, uniform convergence holds.
Exercises
1.2.1.LetGbe a locally compact group and let f,ginL1(G)be supported in the subsetsA andBof G, respectively. Prove that f∗g is supported in the algebraic product setAB.
1.2.2.For a function f on a locally compact groupGandt∈G, lettf(x) = f(tx) and ft(x) =f(xt). Show that
tf∗g=t(f∗g) and f∗gt= (f∗g)t whenever f,g∈L1(G), equipped with left Haar measure.
1.2.3.LetGbe a locally compact group with left Haar measure. Let f ∈Lp(G) and ˜g ∈Lp0(G), where 1< p<∞; recall that g(x) =e g(x−1). For t,x∈G, let
tg(x) =g(tx). Show that for any ε>0 there exists a relatively compact symmet- ric neighborhood of the originU such thatu∈U implies
ueg−eg
Lp0(G)<ε and therefore
|(f∗g)(v)−(f∗g)(w)|<
f Lpε whenevervw−1∈U.
1.2.4.LetGbe a locally compact group and let 1≤p≤∞. Let f ∈Lp(G)andµ be a finite Borel measure onGwith total variation
µ . Define
(µ∗f)(x) = Z
G
f(y−1x)dµ(y).
Show that ifµis an absolutely continuous measure, then the preceding definition extends (1.2.4). Prove that
µ∗f
Lp(G)≤ µ
f
Lp(G).
1.2.5.Show that a Haar measureλ for the multiplicative group of all positive real numbers is
λ(A) = Z ∞
0
χA(t)dt t .
1.2.6.LetG=R2\ {(0,y): y∈R}with group operation(x,y)(z,w) = (xz,xw+y).
[Think ofGas the group of all 2×2 matrices with bottom row(0,1)and nonzero top left entry.] Show that a left Haar measure onGis
λ(A) = Z +∞
−∞
Z +∞
−∞ χA(x,y)dx dy x2 , while a right Haar measure onGis
ρ(A) = Z +∞
−∞
Z +∞
−∞ χA(x,y)dx dy
|x| .
1.2.7.(Hardy [118], [119]) Use Theorem 1.2.10 to prove that Z ∞
0
1 x
Z x 0
|f(t)|dt p
dx 1p
≤ p p−1
f
Lp(0,∞), Z ∞
0
Z ∞ x
|f(t)|dt p
dx 1p
≤p Z ∞
0
|f(t)|ptpdt 1p
, when 1<p<∞.
Hint:On the multiplicative group(R+,dtt )consider the convolution of the function
|f(x)|x1p with the functionx−
1 p0
χ[1,∞)and the convolution of the function|f(x)|x1+1p withx1pχ(0,1].
1.2.8.(G. H. Hardy) Let 0<b<∞and 1≤p<∞. Prove that Z ∞
0
Z x
0
|f(t)|dt p
x−b−1dx 1p
≤ p b
Z ∞
0
|f(t)|ptp−b−1dt 1p
,
Z ∞
0
Z ∞
x
|f(t)|dt p
xb−1dx 1p
≤ p b
Z ∞
0
|f(t)|ptp+b−1dt 1p
.
Hint:On the multiplicative group(R+,dtt )consider the convolution of the function
|f(x)|x1−bp withx−bpχ[1,∞)and of the function|f(x)|x1+bp withxbpχ(0,1].
1.2.9.OnRnletT(f) = f∗K, whereKis a positive L1function and f is inLp, 1≤p≤∞. Prove that the operator norm ofT:Lp→Lpis equal to
K L1. Hint:Clearly,kTkLp→Lp≤ kKkL1. Conversely, fix 0<ε<1 and letNbe a positive integer. LetχN=χB(0,N)and for anyR>0 letKR=KχB(0,R), whereB(x,R)is the ball of radiusRcentered atx. Observe that for|x| ≤(1−ε)N, we haveB(0,Nε)⊆ B(x,N); thusRRnχN(x−y)KNε(y)dy=RRnKNε(y)dy=
KNε
L1. Then K∗χN
p Lp
kχNkLpp
≥
KNε∗χN
p
Lp(B(0,(1−ε)N)
χN
p Lp
≥ KNε
p
L1(1−ε)n. LetN→∞first and thenε→0.
1.2.10.On the multiplicative group(R+,dtt)letT(f) = f∗K, whereKis a positive L1function and f is inLp, 1≤p≤∞. Prove that the operator norm ofT:Lp→Lp is equal to theL1norm ofK. Deduce that the constantsp/(p−1)andp/bare sharp in Exercises 1.2.7 and 1.2.8.
Hint:Adapt the idea of Exercise 1.2.9 to this setting.
1.2.11.LetQj(t) =cj(1−t2)jfort∈[−1,1]and zero elsewhere, wherecjis cho- sen such thatR−11 Qj(t)dt=1 for all j=1,2, . . ..
(a) Show thatcj<√ j.
(b) Use part (a) to show that{Qj}jis an approximate identity onRas j→∞.
(c) Given a continuous function f onRthat vanishes outside the interval[−1,1], show that f∗Qjconverges tof uniformly on[−1,1].
(d) (Weierstrass) Prove that every continuous function on[−1,1]can be approxi- mated uniformly by polynomials.
Hint:Part (a): Consider the integralR|t|≤n−1/2Qj(t)dt. Part (d): Consider the func- tiong(t) =f(t)−f(−1)−t+12 (f(1)−f(−1)).
1.2.12.(Christ and Grafakos [51]) LetF ≥0,G≥0 be measurable functions on the sphereSn−1and letK≥0 be a measurable function on[−1,1]. Prove that
Z
Sn−1 Z
Sn−1
F(θ)G(ϕ)K(θ·ϕ)dϕdθ≤C F
Lp(Sn−1)
G
Lp0(Sn−1),
where 1≤p≤∞,θ·ϕ=∑nj=1θjϕjandC=RSn−1K(θ·ϕ)dϕ,which is independent ofθ. Moreover, show thatCis the best possible constant in the preceding inequality.
Using duality, compute the norm of the linear operator F(θ)7→
Z
Sn−1
F(θ)K(θ·ϕ)dϕ fromLp(Sn−1)to itself.
Hint:Observe thatRSn−1R
Sn−1F(θ)G(ϕ)K(θ·ϕ)dϕdθ is bounded by the quan- tityR
Sn−1
R
Sn−1F(θ)K(θ·ϕ)dθp
dϕ
1
pkGk
Lp0(Sn−1).Apply H¨older’s inequality to the functions F and 1 with respect to the measure K(θ·ϕ)dθ to deduce that R
Sn−1F(θ)K(θ·ϕ)dθ is controlled by Z
Sn−1
F(θ)pK(θ·ϕ)dθ 1/pZ
Sn−1
K(θ·ϕ)dθ 1/p0
.
Use Fubini’s theorem to bound the latter by kFkLp(Sn−1)kGkLp0
(Sn−1) Z
Sn−1
K(θ·ϕ)dϕ.
Note that equality is attained if and only if bothFandGare constants.