Boundedness of L^{1} spectral multipliers for an exponential solvable Lie group

by Waldemar Hebisch^{1}

Introduction

Let M be a measure space and let L be a positive definite operator on L^{2}(M). By the spectral theorem, for any bounded Borel measurable function m : [0, ∞) ↦ ℂ the operator m(L)f = ∫_{0}^{∞}m(λ)dE(λ)f is bounded on L^{2}(M).

We are interested in sufficient conditions on m for m(L) to be bounded on L^{p}(M), p ≠ 2.

A very general theorem of Stein [18] and its strengthening by Cowling [8] give such conditions under the assumption that L generates a semigroup of contractions on L^{1}. Then, indeed, m(L) is bounded on L^{p}(M), 1 < p < ∞, if m is of the form m(λ) = λ ∫_{0}^{∞}e^{−λξ}F (ξ)dξ with F ∈ L^{∞}(ℝ^{+}). The assumption on L is satisfied by many examples, but the condition on m implies that m is holomorphic in the half-plane | arg(λ)| < π/2.

In the case of a Riemmanian manifold of exponential volume growth of riemannian balls (or a noncompact semisimple Lie group) stronger positive results have been proved. However, it is usually assumed that m is holomorphic in a parabolic region around the real half-axis and in some cases it is known that this assumption is necessary [2], [3], [7], [19].

In contrast to the above, what follows from a 1960 result by Hörmander [13] is that if M = ℝ^{n} and L is the (minus) Laplace operator on ℝ^{n}, then a bound on only a finite number of derivatives of m suffices for m(L) to be bounded on L^{p}(M), 1 < p < ∞. This has been generalized to M being a group of polynomial growth [1], or compact manifolds [6], L being a Laplacian. In this work polynomial growth was crucial in the proof and the growth rate determined the required regularity of m.

It turns out however, that the relation of the volume growth of the balls to the regularity of m which guarantees boundedness of m(L) on L^{p} is not straightforward. For some nilpotent Lie groups regularity of m is related to the topological dimension of the group

1 Praca wykonana w ramach projektu badawczego 2 P301 052 07. rather than to its growth [11], [12], [17]. Also on groups NA in the Iwasawa decomposition of a semi-simple Lie group which are of exponential growth for some specific elliptic L, estimates on only a finite number of derivatives of m imply boundedness of m(L) on L^{p}, p ≠ 2, [9], [10]. As these groups and the distinguished Laplacean on them are pretty special, we would like to present another example, this time of a unimodular group of exponential growth and an elliptic Laplacian L on it for which m ∈ C_{c}^{6}(ℝ^{+}) implies that m(L) is bounded on all L^{p}, 1 ≤ p < ∞.

Our methods allow to handle a bit more general groups but, as they are now, seem to be too crude to solve the problem in general. In fact, Christ and Müller [5] give an example of a solvable Lie group on which m must be holomorphic. There is a large disparity between our methods and the ones of [5]. This is the reason why some arguments are presented in a more general form then actually needed for the proof here. Also in Lemmas (1.1) and (1.2) we give sharp estimates for the constants involved, as those might be of interest of their own.

Preliminaries

Let G be equal to ℝ^{3}, the multiplication being given by the formula

(t_{1}, x_{1}, y_{1})(t_{2}, x_{2}, y_{2}) = (t_{1} + t_{2}, e^{t_{2}}x_{1} + x_{2}, e^{−t_{2}}y_{1} + y_{2}).

We consider left invariant vector fields X, Y, T

X = ∂_{x},  Y = ∂_{y},  T = ∂_{t} + x∂_{x} − y∂_{y}.

The corresponding right invariant vector fields X̃, Ỹ, T̃ are

X̃ = e^{t}∂_{x},  Ỹ = e^{−t}∂_{y},  T̃ = ∂_{t}.

Let

L = X^{2} + Y^{2} + T^{2},  L̃ = X̃^{2} + Ỹ^{2} + T̃^{2}

and let p_{t} be the convolution kernel of exp(tL):

exp(tL)f = f ∗ p_{t}.

We need the formula

exp(tL̃)f = p_{t} ∗ f.

This formula is valid for all sublaplaceans on Lie groups. For the proof one may consider the distribution K_{L} supported at identity such that

Lf = f ∗ K_{L}.

Then

L̃f = K_{L} ∗ f

and

∂_{t}(f ∗ p_{t}) = ∂_{t} exp(tL)f = L exp(tL)f = exp(tL)Lf = f ∗ K_{L} ∗ p_{t} = f ∗ (L̃p_{t}).

So

∂_{t}p_{t} = L̃p_{t}

and

∂_{t}(p_{t} ∗ f) = (∂_{t}p_{t}) ∗ f = (L̃p_{t}) ∗ f = L̃(p_{t} ∗ f).

As this equation has a unique solution for f in the domain of L̃, we have p_{t} ∗f = exp(tL̃)f. Schrödinger operators As we shall see toward the end of the paper, our considerations here cannot be re-

stricted to sublaplaceans. We have to consider also Schrödinger operators that is operators of the form

H = ∑ U_{j}^{∗}U_{j}

where U_{j} = X_{j}+iV_{j}, X_{j} are vector fields and V_{j} are locally square integrable real functions. Our operators U_{j} are closed operators with a common core C_{c}^{∞}(G). Consequently, by the von Neumann theorem (cf. for example [20] Theorem 5.39) H is selfadjoint.

From now on we will assume that X_{j} generate G. For the proof of main theorem of the paper we assume even more. Then it is assumed that X_{j} form a linear basis of the Lie algebra of G. Let d(x) be the optimal control distance from x ∈ G to e associated to the X_{j}’s.

(1.1). Lemma. For every f in C_{c}^{∞}(G) and γ > 0 we have: γ|ℑ(Hf, e^{2sd}f)| ≤ ℜ(Hf, e^{2sd}f) + s^{2}(1 + γ)^{2}∥e^{sd}f∥^{2}.

Proof: We have

(U_{j}^{∗}U_{j}f, e^{2sd}f) = (U_{j}f, U_{j}(e^{2sd}f)) = (U_{j}f, e^{2sd}U_{j}f) + (U_{j}f, 2sX_{j}(d)e^{2sd}f) =

∥e^{sd}U_{j}f∥^{2} + (U_{j}f, 2sX_{j}(d)e^{2sd}f)

and

|((U_{j}^{∗}U_{j}f, e^{2sd}f) − ∥e^{sd}U_{j}f∥^{2}| = |(U_{j}f, 2sX_{j}(d)e^{2sd}f)| ≤ 2s∥e^{sd}U_{j}f∥∥X_{j}(d)e^{sd}f∥

≤ α∥e^{sd}U_{j}f∥^{2} + s^{2}/α∥X_{j}(d)e^{sd}f∥^{2}.

Next, by the inequality ∑ |X_{j}d(x)|^{2} ≤ 1,

∑ ∥(X_{j}d)e^{sd}f∥^{2} ≤ ∥e^{sd}f∥^{2}

so

ℜ(Hf, e^{2sd}f) + s^{2}/α∥e^{sd}f∥^{2} ≥ (1 − α) ∑ ∥e^{sd}U_{j}f∥^{2}

and

|ℑ(Hf, e^{2sd}f)| ≤ α ∑ ∥e^{sd}U_{j}f∥^{2} + s^{2}/α∥e^{sd}f∥^{2}.

Putting α = (1 + γ)^{−1} we have:

γ|ℑ(Hf, e^{2sd}f)| ≤ γ/(1 + γ) ∑ ∥e^{sd}U_{j}f∥^{2} + γ(1 + γ)s^{2}∥e^{sd}f∥^{2}

= (1 − α) ∑ ∥e^{sd}U_{j}f∥^{2} + γ(1 + γ)s^{2}∥e^{sd}f∥^{2}
≤ ℜ(Hf, e^{2sd}f) + ((1 + γ) + γ(1 + γ))s^{2}∥e^{sd}f∥^{2}

which is our conclusion.

For a real s let

(f, g)_{s} = ∫_{G} f(x)ḡ(x)e^{2sd(x)}dx

be the inner product in L^{2}(e^{2sd}). We write ∥f∥_{s} for the corresponding norm.

(1.2). Lemma. Let H and d be as above. For s negative (or positive) the semigroup e^{−zH}, ℜz > 0 extends (restricts) to a holomorphic semigroup on the weighted space L^{2}(e^{2sd}). Moreover,

∥e^{−zH}∥_{L^{2}(e^{2sd}),L^{2}(e^{2sd})} ≤ exp(ℜz(1 + |\frac{ℑz}{ℜz} |)^{2}s^{2}).

Proof: This is a consequence of (1.1) . Indeed,we rewrite (1.1) as

γ|ℑ(Hf, f)_{s}| ≤ ℜ(Hf, f)_{s} + C∥f∥_{s}^{2}.

where C = s^{2}(1 + γ)^{2}. In general f ranges over C_{c}^{∞}(G), but if s ≤ 0 we may extend the inequality above to the domain of H. To see this we write

⟨f, g⟩_{C,s} = \frac{1}{2}(((H + CI)f, g)_{s} + (f, (H + CI)g)_{s}).

⟨f, g⟩_{C,s} is a new scalar product on a dense subspace of L^{2}(e^{2sd}) and our inequality extends to the domain of the corresponding quadratic form. From the proof of (1.1) we have

⟨f, f⟩_{C,s} = ℜ(Hf, f)_{s} + C∥f∥_{s}^{2} ≈ ∑ ∥e^{sd}U_{j}f∥^{2} + C∥f∥_{s}^{2}

so ⟨f, f⟩_{C,s} is finite for f in the domain of H. We then have

ℜ(z(H + CI)f, f) ≥ 0

for f in the domain of H and z in the sector S_{γ} = {z : |ℑz| ≤ γℜz}. This means that the operator A = −z(H + CI) is dissipative in the L^{2}(e^{2sd})-norm. Moreover,

e^{tA} = lim_{n→∞}(I − \frac{t}{n}A)^{−n}.

On the right hand side we have a sequence of contractions on L^{2}(e^{2sd}) which is convergent on functions from L^{2}, which is a dense subset of L^{2}(e^{2sd}), s < 0. Therefore e^{tA} extends to contractions on L^{2}(e^{sd}). This completes the proof of the lemma for negative s. For s > 0 we obtain the lemma by duality.

(1.3). Lemma. If t > 0 and f ∈ L^{2}

|e^{−tH}f| ≤ exp(−t ∑ X_{j}^{∗}X_{j})|f|

almost everywhere.

Remark The Lemma is valid in general, but we will prove it only when {exp(tX_{j}) : t ∈ ℝ} is unbounded. For example, if G is exponential. Then for every locally integrable function V_{j} there exists a Borel function F_{j} such that X_{j}F_{j} = V_{j}.

Proof: By the Trotter formula [15] it is enough to prove the lemma for a single U = X + iV. By assumption, there is a F such that XF = V. Then

exp(−iF)X exp(iF) = X + iV = U, exp(−iF)X^{∗} exp(iF) = X^{∗} − iV = U^{∗}

and

exp(−iF)X^{∗}X exp(iF) = U^{∗}U

so

exp(−iF) exp(−tX^{∗}X) exp(iF) = exp(−tU^{∗}U).

Since exp(−iF) has absolute value 1 the Lemma follows.

In the sequel we will denote the kernel of the semigroup by e^{−zH}δ_{e}, the second argument being replaced by e.

(1.4). Lemma. There is a constant C independent of V_{j} such that for all real l and s

∥e^{−(1+il)H}δ_{e}∥_{L^{2}(e^{sd})} ≤ C exp(C(1 + l^{2})s^{2})

Proof: Note that (1.3) implies |e^{−(1/2)H}δ_{e}| ≤ exp(−(1/2) ∑ X_{j}^{∗}X_{j}) = p_{1/2}. As for real z and p_{z} the estimate is known, the Lemma follows from (1.2) .

Main Theorems

Now we come back to our group G as defined in the preliminaries. Let d be (left) invariant riemannian metric on G. There is a constant C such that

B_{r} = {(t, x, y) : d((t, x, y), 0) ≤ r} ⊂

{(t, x, y) : |t| < r, |x| < C(e^{r} + 1), |y| < C(e^{r} + 1)}.

Let a weight function w be defined as w(t, x, y) = |xy|. A straightforward calculation shows that for some C

∫_{d(g,0)<r} (1 + w(g))^{−1}dg ≤ Cr^{3}.

(1.5). Theorem. There exists C such that for every s ∈ ℝ we have

∥p_{1+is}∥_{L^{1}(G)} ≤ C(1 + |s|^{5}).

(1.6). Theorem. For every compactly supported F ∈ C^{6} (or F in the Sobolev space H^{( \frac{11}{2} +ϵ)}) the operator F (−L) is bounded on L^{1}(G)

Theorem (1.6) is a consequence of (1.5) . Indeed, using the spectral theorem and the inversion formula for the Fourier transform we have

F (−L) = \frac{1}{2π} ∫ f̂(s)p_{1−is}ds,

where f(x) = F (x)e^{x}. F ∈ H^{( \frac{11}{2} +ϵ)} implies that ∫ |f̂(s)|(1 + |s|)^{5}ds is convergent. We are going to prove (1.5) . From (1.4) (putting V_{j} = 0) we know that

∫ |p_{1+is}(g)|e^{d(g,e)}dg
≤ (∫ |p_{1+is}(g)|^{2}e^{(C+1)d(g,e)}dg)^{1/2} (∫ e^{−2Cd(x,e)}dg)^{1/2}
≤ C exp(Cs^{2}).

Consequently, if r = Cs^{2}

∫ |p_{1+is}(g)|dg ≤ ∫_{d(g,0)<r} |p_{1+is}(g)|dg + ∫_{d(g,0)≥r} |p_{1+is}(g)|dg

≤ ∫_{d(g,0)<r} |p_{1+is}(g)(1 + w(g))^{1/2}|(1 + w(g))^{−1/2}dg + e^{−r} ∫_{d(g,0)≥r} |p_{1+is}(g)|e^{d(g,0)}dg

≤ ∥p_{1+is}(1 + w)^{1/2}∥_{L^{2}}∥(1 + w)^{−1/2}∥_{L^{2}(B_{r})} + e^{−r} ∫ |p_{1+is}(g)|e^{d(g,0)}dg
≤ C(1 + |s|^{2})^{3/2}∥p_{1+is}(1 + w)^{1/2}∥_{L^{2}} + e^{−Cs^{2}}Ce^{Cs^{2}}.

We are going to show that

(1.7)  ∥p_{1+is}(1 + w)^{1/2}∥_{L^{2}}^{2} ≤ C(1 + |s|)^{4}

and

(1.8)  |∂_{s}∥p_{1+is}w^{1/2}∥_{L^{2}}^{2}| ≤ C(1 + |s|)^{3}.

We note that w ≤ Ce^{cd} for some C and c because w ≤ (1+|x|)(1+|y|) and and (x, y, u) → (1 + |x|)(1 + |y|) is submultiplicative, by [14]. Consequently, the left hand side of both (1.7) and (1.8) is finite, and (1.8) implies (1.7). To prove (1.8) we write

|∂_{s}∥p_{1+is}w^{1/2}∥_{L^{2}}^{2}| = 2|ℜ⟨iLp_{1+is}, wp_{1+is}⟩| = 2|ℑ(⟨Xp_{1+is}, X(wp_{1+is})⟩+

⟨Y p_{1+is}, Y (wp_{1+is})⟩ + ⟨T p_{1+is}, T (wp_{1+is})⟩)|.

Since T w = 0, we have

2|ℑ(⟨Xp_{1+is}, X(wp_{1+is})⟩ + ⟨Y p_{1+is}, Y (wp_{1+is})⟩ + ⟨T p_{1+is}, T (wp_{1+is})⟩)| =
2|ℑ(⟨Xp_{1+is}, (Xw)p_{1+is}⟩ + ⟨Y p_{1+is}, (Y w)p_{1+is}⟩)| ≤
2(|⟨Xp_{1+is}, sgn(x)|y|p_{1+is}⟩| + |⟨Y p_{1+is}, sgn(y)|x|p_{1+is}⟩|).

We are going to estimate the second term, the argument for the first being similar. We write

|⟨Y p_{1+is}, sgn(y)|x|p_{1+is}⟩| ≤ ∥|x|^{1/2}p_{1+is}∥_{L^{2}}∥|x|^{1/2}Y p_{1+is}∥_{L^{2}}.

Now we note that

∥|x|^{1/2}Y p_{1+is}∥_{L^{2}} ≤ C( sup_{|z−1|<1/2} ∥|x|^{1/2}p_{z+is}∥_{L^{2}} + 1).

Indeed,

−⟨Lp_{1+is}, |x|p_{1+is}⟩ = ∥|x|^{1/2}T p_{1+is}∥_{L^{2}}^{2} + ∥|x|^{1/2}Xp_{1+is}∥_{L^{2}}^{2} + ∥|x|^{1/2}Y p_{1+is}∥_{L^{2}}^{2}+

⟨T p_{1+is}, |x|p_{1+is}⟩ + ⟨Xp_{1+is}, sgn(x)p_{1+is}⟩ ≥

∥|x|^{1/2}Y p_{1+is}∥_{L^{2}}^{2} + ∥|x|^{1/2}T p_{1+is}∥_{L^{2}}^{2} − ∥|x|^{1/2}T p_{1+is}∥_{L^{2}}∥|x|^{1/2}p_{1+is}∥_{L^{2}}−

∥Xp_{1+is}∥_{L^{2}}∥p_{1+is}∥_{L^{2}} ≥

∥|x|^{1/2}Y p_{1+is}∥_{L^{2}}^{2} − \frac{1}{2}∥|x|^{1/2}p_{1+is}∥_{L^{2}}^{2} − C

but since p_{z} depends analytically on z we have

∥|x|^{1/2}Lp_{1+is}∥_{L^{2}} = ∥|x|^{1/2}∂_{s}p_{1+is}∥_{L^{2}} ≤ 2 sup_{|z−1|<1/2} ∥|x|^{1/2}p_{z+is}∥_{L^{2}}.

As for the z’s |z − 1| < \frac{1}{2} the estimate trivially reduces to the case z = 1, we consider only this case.

Thus we end up with the task to prove

∥|x|^{1/2}p_{1+is}∥_{L^{2}}^{2} ≤ C(1 + |s|^{3}).

We have

|∂_{s}∥|x|^{1/2}p_{1+is}∥_{L^{2}}^{2}| = |2ℜ⟨iL̃p_{1+is}, |x|p_{1+is}⟩| =
|2ℑ⟨X̃^{2}p_{1+is}, |x|p_{1+is}⟩| = |2ℑ⟨e^{2t}∂_{x}p_{1+is}, sgn(x)p_{1+is}⟩| =
|2⟨e^{t3/2}∂_{x}p_{1+is}, sgn(x)e^{t/2}p_{1+is}⟩| ≤ 2∥e^{t/2}X̃p_{1+is}∥_{L^{2}}∥e^{t/2}p_{1+is}∥_{L^{2}}.

We are going to estimate the second factor, the estimate for the first is similar.

We decompose the left regular representation of G using the Fourier transform in x coordinate. Put

H_{x} = −∂_{t}^{2} − e^{−2t}∂_{y}^{2} + x^{2}e^{2t}.

H_{x} is a Schrödinger operator on two dimensional solvable group G_{0}. We have

∥e^{t/2}p_{1+is}∥_{L^{2}}^{2} = ∫_{−∞}^{+∞}∥e^{t/2} exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2}dx =

∫_{|x|<e^{−c|s|^{2}}} + ∫_{e^{−c|s|^{2}} ≤|x|<1} + ∫_{|x|≥1}

By (1.4)

∥e^{t/2} exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}}^{2} ≤ Ce^{Cs^{2}}

independently of x ∈ ℝ. Hence the first of the three integrals above is bounded by a constant provided that c ≥ C.

Since L̃ is elliptic on G we have

∥X̃^{2}f∥_{L^{2}(G)} ≤ C(∥L̃f∥_{L^{2}(G)} + ∥f∥_{L^{2}(G)}).

This inequality remains valid in every unitary representation, so

∥x^{2}e^{2t}f∥_{L^{2}(G_{0})} ≤ C(∥H_{x}f∥_{L^{2}(G_{0})} + ∥f∥_{L^{2}(G_{0})})

and

∥e^{t/2}f∥_{L^{2}(G_{0})} ≤ ∥e^{2t}f∥_{L^{2}(G_{0})}^{1/4}∥f∥_{L^{2}(G_{0})}^{3/4} = |x|^{−1/2}∥x^{2}e^{2t}f∥_{L^{2}(G_{0})}^{1/4}∥f∥_{L^{2}(G_{0})}^{3/4} ≤

C|x|^{−1/2}(∥H_{x}f∥_{L^{2}(G_{0})} + ∥f∥_{L^{2}(G_{0})})^{1/4}∥f∥_{L^{2}(G_{0})}^{3/4}.

We have

∥e^{t/2} exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2} ≤

C|x|^{−1}(∥H_{x} exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})} + ∥ exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})})^{1/2}×

∥ exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{3/2} ≤

C|x|^{−1}(∥H_{x} exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2} + ∥ exp(−(1 + is)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2}) =

C|x|^{−1}(∥H_{x} exp(−H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2} + ∥ exp(−H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2}) ≤
C|x|^{−1}∥ exp(−(1/2)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2}

so the second integral is bounded by 2C ∫_{exp(−c|s|^{2})}^{1} x^{−1}dx = C_{1}|s|^{2}.

If |x| ≥ 1, then

∫_{|x|≥1} \frac{1}{|x|}∥ exp(−(1/2)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2} ≤ ∫_{−∞}^{+∞}∥ exp(−(1/2)H_{x})δ_{0}∥_{L^{2}(G_{0})}^{2} = ∥p_{1/2}∥_{L^{2}(G)}^{2}

so the third integral is bounded by a constant. Consequently

∥e^{t/2}p_{1+is}∥_{L^{2}}^{2} ≤ C(1 + |s|^{2})

which ends the proof.

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