Time decay for the bounded mean oscillation of solutions of the Schrödinger and wave equations

S.J. Montgomery-Smith* University of Missouri

Abstract: Let u(x, t) be the solution of the Schrödinger or wave equation with L_{2} initial data. We provide counterexamples to plausible conjectures involving the decay in t of the BMO norm of u(t, ·). The proofs make use of random methods, in particular, Brownian motion.

1. Introduction Consider the wave equation:

∂_{t}^{2}u(t, x) = Δu(t, x)
u(0, x) = 0
∂_{t}u(0, x) = f,

where t ∈ R, x ∈ R^{d}, and f ∈ L_{2}(R^{d}). Let us write once and for all B_{t}f = u(t, ·), so that

B̂_{t}f(ζ) = \frac{sin(t |ζ|)}{|ζ|}f̂(ζ). In this paper we will be discussing the endpoints of the following

assertion. (B_{d,p,q}): if f ∈ L_{2}(R^{d}), then t ↦ B_{t}f is in L_{q}(R, L_{p}(R^{d})), and there exists a constant c independent of f such that

(∫_{−∞}^{∞} ∥B_{t}f∥_{p}^{q} dt)^{1/q} ≤ c ∥f∥_{2} .

If p, q ≥ 1, then by standard arguments, it is easy to show that this is equivalent to its dual assertion. Here, as in the rest of the paper, 1/p + 1/p′ = 1/q + 1/q′ = 1. (B_{d,p′,q′}^{∗}): if t ↦ f_{t} is in L_{q′}(R, L_{p′}(R^{d})), then ∫_{−∞}^{∞} B_{t}f_{t} dt (exists almost everywhere and) is in L_{2}(R^{d}), and there exists a constant c independent of f_{t} such that

∥∫_{−∞}^{∞} B_{t}f_{t} dt∥_{2} ≤ c (∫_{−∞}^{∞} ∥f_{t}∥_{p′}^{q′} dt)^{1/q′} .

In stating these assertions, we will always suppose that d, p and q satisfy the following conditions:

\begin{matrix}\frac{d}{p} + \frac{1}{q} & = \frac{d}{2} − 1 \\ p, q & ≥ 2 \\ d & ≥ 3. \end{matrix} \}  (∗_{B})

* Research funded in part by N.S.F. D.M.S. 9424396.

Indeed the first condition can easily be shown to be necessary if (B_{d,p,q}) is to hold. This follows from considering the substitution f(x) → f(rx) for any 0 < r < ∞. (Physicists might call this a dimensional argument.)

It is known that if conditions (∗_{B}) hold, then assertions (B_{d,p,q}) and (B_{d,p′,q′}^{∗}) are true whenever p < ∞. Results of this form are commonly known as Strichartz inequalities, and were proven in [St1] in the case p = q. The other cases follow by the same argument. The main application of these results is to show existence and uniqueness results for non-linear wave equations. (See [Br], [Ka], [Ke] and [Ru] for applications of this and similar results, and references to other such results.)

The only case left is p = ∞, q = 2 and d = 3. The assertion (B_{3,∞,2}) and its dual (B_{3,1,2}^{∗}) were shown to be false by Klainerman and Machedon [Kl].

It was then conjectured that a weaker assertion may hold, that is, (B_{3,BMO,2}). The space BMO(R^{d}) has long been studied by many authors. Many results that are not true for L_{∞} turn out to be true for BMO. The space BMO is slightly larger than L_{∞}, with a smaller norm. There are several equivalent definitions for this space (see for example [Se]), and for our purposes it will be convenient to define BMO in terms of its dual space, H_{1}, that is, BMO(R^{d}) is the space of measurable functions f from R^{d} such that if g ∈ H_{1}(R^{d}), then fg ∈ L_{1}(R^{d}). Furthermore, BMO is equipped with the dual norm:

∥f∥_{BMO} = sup {∫_{R^{d}} fg : ∥g∥_{H_{1}} ≤ 1} .

Thus it only remains to define H_{1}(R^{d}). Again, the literature gives several definitions (see for example [Se]). We will pick a definition in terms of maximal functions based upon the heat kernel:

Mg(x) = sup_{t>0} ∫_{R^{d}} \frac{1}{(4πt)^{d/2}} exp(− |y|^{2} /t)g(x − y) dy.

Then we say that g ∈ H_{1}(R^{d}) if Mg ∈ L_{1}(R^{d}), and set

∥g∥_{H_{1}} = ∥Mg∥_{1} .

It is clear that |g| ≤ |Mg| almost everywhere, and hence H_{1} is a subspace of L_{1} with larger norm.

For definiteness, let us explicitly state the assertions involving these norms. (B_{3,BMO,2}): if f ∈ L_{2}(R^{3}), then t ↦ B_{t}f is in L_{2}(R, BMO(R^{3})), and there exists a constant c independent of f such that

(∫_{−∞}^{∞} ∥B_{t}f∥_{BMO}^{2} dt)^{1/2} ≤ c ∥f∥_{2} .

This is equivalent to its dual assertion: (B_{3,H_{1},2}^{∗}): if t ↦ f_{t} is in L_{2}(R, H_{1}(R^{3})), then ∫_{−∞}^{∞} B_{t}f_{t} dt is in L_{2}(R^{3}), and there exists a constant c independent of f_{t} such that

∥∫_{−∞}^{∞} B_{t}f_{t} dt∥_{2} ≤ c (∫_{−∞}^{∞} ∥f_{t}∥_{H_{1}}^{2} dt)^{1/2} .

The purpose of this paper is to provide a counterexample to assertion (B_{3,BMO,2}). We will proceed by considering the dual assertion (B_{3,H_{1},2}^{∗}). We will be using random methods, and so we do not provide an explicit counterexample. The tool from probability theory we shall use is Brownian motion. We refer the reader to [Pe] for details. However the only property of Brownian motion we shall use is that it is a randomly chosen continuous function t ↦ b_{t} from R to R such that b_{t} − b_{s} is a gaussian random variable with mean 0 and standard deviation \sqrt{|s − t|}. Throughout the paper, we will use the notation EX to denote the expected value of a random variable X.

We shall also consider the Schrödinger equation with zero potential:

∂_{t}u(t, x) = −iΔu(t, x)
u(0, x) = f(x),

where, once again, t ∈ R, x ∈ R^{d}, and f ∈ L_{2}(R^{d}). Let us write once and for all A_{t}f = u(t, ·), so that Â_{t}f(ζ) = exp(i |ζ|^{2})f̂(ζ). Then we get a similar assertion, with its dual. (A_{d,p,q}): if f ∈ L_{2}(R^{d}), then t ↦ A_{t}f is in L_{q}(R, L_{p}(R^{d})), and there exists a constant c independent of f such that

(∫_{−∞}^{∞} ∥A_{t}f∥_{p}^{q} dt)^{1/q} ≤ c ∥f∥_{2} .

(A_{d,p′,q′}^{∗}): if t ↦ f_{t} is in L_{q′}(R, L_{p′}(R^{d})), then ∫_{−∞}^{∞} A_{t}f_{t} dt is in L_{2}(R^{d}), and there exists a constant c independent of f_{t} such that

∥∫_{−∞}^{∞} A_{t}f_{t} dt∥_{2} ≤ c (∫_{−∞}^{∞} ∥f_{t}∥_{p′}^{q′} dt)^{1/q′} .

In stating these assertions, we will always suppose that p, q and d satisfy the following conditions:

\begin{matrix}\frac{d}{p} + \frac{2}{q} & = \frac{d}{2} \\ p, q & ≥ 2 \\ d & ≥ 2. \end{matrix} \}  (∗_{A})

Once again, the first condition can be shown to be necessary.

Rather less is known about these assertions than the corresponding ones for the wave equation. It is known [Gi] that if conditions (∗_{A}) hold, and q > 2, then (A_{d,p,q}) holds.

However the case q = 2 seems to be open. This paper tackles one of these problems, that is, the case when q = 2, d = 2 and p = ∞. We will demonstrate that the assertion (A_{2,∞,2}) does not hold. (We will also include an alternative proof of this fact due to Tony Carbery and Steve Hofmann.) The problem when q = 2 and d ≥ 3 seems to be very difficult, and at the time of writing is apparently unknown.

We will also deal with the assertions involving the space BMO, again showing that these are false. (A_{2,BMO,2}): if f ∈ L_{2}(R^{2}), then t ↦ A_{t}f is in L_{2}(R, BMO(R^{2})), and there exists a constant c independent of f such that

(∫_{−∞}^{∞} ∥A_{t}f∥_{BMO}^{2} dt)^{1/2} ≤ c ∥f∥_{2} .

This is equivalent to its dual assertion: (A_{2,H_{1},2}^{∗}): if t ↦ f_{t} is in L_{2}(R, H_{1}(R^{2})), then ∫_{−∞}^{∞} A_{t}f_{t} dt is in L_{2}(R^{2}), and there exists a constant c independent of f_{t} such that

∥∫_{−∞}^{∞} A_{t}f_{t} dt∥_{2} ≤ c (∫_{−∞}^{∞} ∥f_{t}∥_{H_{1}}^{2} dt)^{1/2} .

The author would like to acknowledge the tremendous help of many colleagues, who explained the problems, and were a source of ideas and inspiration. In particular, he would like to mention Tony Carbery, Loukas Grafakos, Steve Hofmann, Nigel Kalton, Lev Kapitansky, and Luis Vega.

2. Solutions of the Schrödinger equation We will start by considering the Schrödinger equation, because the techniques are simpler. Our first result will be eclipsed by Theorem 2, given later. However, we will prove the following result because the method of the proof is simpler, and illustrates the main ideas that will be used. Because this will not be a definitive result, we will not be completely rigorous.

Throughout this section, we will make great use of the fact that if f̂(ζ) = exp(−α |ζ|^{2}) (ζ ∈ R^{2}), where α is a complex number with Re(α) > 0, then f(x) = (4πα)^{−1} exp(− |x|^{2} /α). If one is considering tempered distributions, the result remains true if Re(α) = 0.

Theorem 1. Assertion (A_{2,∞,2}) is not true.

In fact, what we will do is to show that assertion (A_{2,1,2}^{∗}) is not true. Our counterexample is f_{t}(x) = α_{t}δ(x−p_{t}), where ∫_{−∞}^{∞} |α_{t}|^{2} dt = 1, p_{t} ∈ R^{2} will be chosen later, and δ is the Dirac delta function on R^{2}. Of course, δ is not a function, and thus is not in L_{1}(R^{2}). However, we will sacrifice rigor for the sake of clarity. A more rigorous argument may be formed by setting δ̂(ζ) = exp(− |ζ|^{2}), and following the argument used in the proof of Theorem 2.

Note that if t ∈ R, and p, q ∈ R^{2}, then

∫_{R^{2}}(A_{t}δ)(x − p)\overline{δ(x − q)} dx = ∫_{R^{2}} exp(it |ζ|^{2}) exp(ip · ζ) exp(−iq · ζ) dζ

= −\frac{1}{4πit} exp (\frac{|p − q|^{2}}{it} ) .

Then

∥∫_{−∞}^{∞} A_{t}f_{t} dt∥_{2}^{2} = ∫_{R^{2}} ∫_{−∞}^{∞} A_{t}f_{t}(x) dt ∫_{−∞}^{∞} \overline{A_{s}f_{s}(x)} ds dx

= ∫_{−∞}^{∞} ∫_{−∞}^{∞} ∫_{R^{2}}(A_{t−s}f_{t})(x)\overline{f_{s}(x)} dx ds dt

= − ∫_{−∞}^{∞} ∫_{−∞}^{∞} \frac{α_{t}ᾱ_{s}}{4πi(t − s)} exp (\frac{|p_{t} − p_{s}|^{2}}{i(t − s)} ) ds dt.

Now we let p_{t} = (\sqrt{θ}b_{t}, 0), where θ is chosen so that if γ is a gaussian random variable with mean 0 and standard deviation 1, then a = E(sin(θγ^{2}) ≠ 0. Thus

E sin (\frac{|p_{t} − p_{s}|^{2}}{(t − s)} ) = E sin(sgn(t − s)θγ^{2}) = a sgn(t − s).

Then, if α_{t} is real,

− Re E ∥∫_{−∞}^{∞} A_{t}f_{t} dt∥_{2}^{2} = ∫_{−∞}^{∞} ∫_{−∞}^{∞} \frac{α_{t}α_{s}}{4πi(t − s)}E sin (\frac{|p_{t} − p_{s}|^{2}}{(t − s)} ) ds dt

= a ∫_{−∞}^{∞} ∫_{−∞}^{∞} \frac{α_{t}α_{s}}{4πi |t − s|} ds dt.

Taking α_{t} = 1 if |t| ≤ 1, and 0 otherwise, we obtain an unbounded integral, and hence the desired counterexample.

We will also give a different proof of this last result due to Tony Carbery and Steve Hofmann. This proof was found shortly after the one just given, and is reproduced here with their permission. Again, we are sacrificing some rigor to obtain clarity. The starting point is the same, except that we shall take f_{t}(x) = δ(x − \sqrt{r}p_{t}) for t ∈ I, and 0 otherwise, where I is any interval. We shall suppose that p_{t} is any path in Lip_{1/2}, with ∥p∥_{Lip_{1/2}} = 1, and that r ∈ R. Then it is sufficient to find a counterexample to the following assertion. For every r ∈ R and interval I we have

|∫_{I} ∫_{I} \frac{1}{4πi(t − s)} exp (\frac{r |p_{t} − p_{s}|^{2}}{i(t − s)} ) ds dt| ≤ |I| .

Now we apply an idea from [Co]. Pick F ∈ C_{0}^{∞}(R) such that F (ρ) = ρ for |ρ| ≤ 1. Then

\frac{|p_{t} − p_{s}|^{2}}{t − s} = ∫_{−∞}^{∞} F̂(−r) exp (\frac{r |p_{t} − p_{s}|^{2}}{i(t − s)} ) dr,

and so combining the last two displayed equations, and rearranging the integrals, we see that

|∫_{I} ∫_{I} \frac{|p_{t} − p_{s}|^{2}}{4π(t − s)^{2}} ds dt| ≤ |I| ∫_{−∞}^{∞}|F̂(r)| dr ≤ C |I| ,

where C is some universal constant. But by a result of Strichartz [St2], we have that if D^{1/2}p ∈ BMO, then

sup_{I} \frac{1}{|I|} ∫_{I} ∫_{I} \frac{|p_{t} − p_{s}|^{2}}{(t − s)^{2}} ds dt ≈ ∥D^{1/2}p∥_{BMO}.

Here D̂^{1/2}p(τ) = |τ|^{1/2} p̂(τ). Hence we have our counterexample by picking p with ∥p∥_{Lip_{1/2}} = 1, and ∥D^{1/2}p∥_{BMO} arbitrarily large.

Now we will improve this result. In the arguments that follow, the reader may feel uncomfortable with the cavalier and implicit use of Fubini’s Theorem. The use of Fubini’s Theorem requires all the integrals to be absolutely convergent, and this is not the case, as it is exactly the opposite that we are trying to show. For this reason, the proofs should really be seen as proofs by contradiction, that is, the reader should suppose initially that the assertions stated in the theorems are true.

Theorem 2. Assertion (A_{2,BMO,2}) is not true.

Once again, we provide a counterexample to the dual assertion (A_{2,H_{1},2}^{∗}). In this case, our counterexample will be f_{t}(x) = α_{t}g(x − p_{t}), where ∫_{−∞}^{∞} |α_{t}|^{2} dt = 1, p_{t} ∈ R^{2} will be chosen later, and ĝ(ζ) = |ζ|^{2} exp(− |ζ|^{2}).

Let us first show that g ∈ H_{1}(R^{2}). Using Fourier transforms, we see that

Mg(x) = sup_{t>0} \frac{1}{4πt} ∫_{R^{2}} exp(− |x − y|^{2} /t)g(y) dy

= sup_{t>0} \frac{4(1 + t) − 2 |x|^{2}}{4π(1 + t)^{3}} exp (−\frac{|x|^{2}}{1 + t})

= O(\frac{1}{1 + |x|^{2}} ),

which is in L_{1}(R^{2}).

Note also that for t ∈ R and p, q ∈ R^{2}

∫R^{2}(A_{t}g)(x − p)\overline{g(x − q)} dx

= ∫_{R^{2}} |ζ|^{4} exp((−2 + it) |ζ|^{2}) exp(ζ · (q − p)) dζ

= \frac{1}{4π(2 − it)}Δ^{2} exp (\frac{|x|^{2}}{−2 + it}) |_{x=q−p}

= \frac{1}{4π(2 − it)^{3}} P (|q − p|^{2} /(−2 + it)) exp (\frac{|q − p|^{2}}{−2 + it})

= \frac{1}{4π(2 − it)^{3}} P (|q − p|^{2} /(−2 + it)) exp (−\frac{2 |q − p|^{2}}{4 + t^{2}} − \frac{it |q − p|^{2}}{4 + t^{2}} ) ,

where P (t) = 16t^{2} − 64t + 32.

Hence

∥∫_{−∞}^{∞} A_{t}f_{t} dt∥_{2}^{2} = ∫_{−∞}^{∞} ∫_{−∞}^{∞} α_{t}ᾱ_{s}K(t, s) ds dt,

where

K(t, s) = \frac{1}{4π(2 − i(t − s))^{3}} P (|p_{t} − p_{s}|^{2} /(−2 + i(t − s)))×

exp (−\frac{2 |p_{t} − p_{s}|^{2}}{4 + (t − s)^{2}} − \frac{i(t − s) |p_{t} − p_{s}|^{2}}{4 + (t − s)^{2}} ) .

Now we will choose p_{t} = (t, \sqrt{θ}b_{t}), where b_{t} and θ are defined as in the previous section. In order to demonstrate that we have a counterexample, we need to show that E(K(t, s)) is not the kernel of bounded operator from L_{2}(R) → L_{2}(R).

We shall be interested in the behavior of E(K(t, s)) as t − s → ±∞. Let b_{t} − b_{s} = \sqrt{|t − s|}γ, where γ is a gaussian random variable with mean 0 and standard deviation 1. Denote ⟨t⟩ = 1 + |t|.

First see that

P (|p_{t} − p_{s}|^{2} /(−2 + i(t − s))) = −16(t − s)^{2} + O(γ^{4}⟨t − s⟩).

Also, there exists a constant c_{1} such that

exp (−\frac{2 |p_{t} − p_{s}|^{2}}{4 + (t − s)^{2}} ) = exp(−(2+O(γ^{2}⟨t−s⟩^{−1})) = exp(−2)+O(exp(c_{1}γ^{2}⟨t−s⟩^{−1})−1).

Furthermore,

\frac{(t − s) |p_{t} − p_{s}|^{2}}{4 + (t − s)^{2}} = (t − s) + sgn(t − s)θγ^{2} + O(γ^{2}⟨t − s⟩^{−1}),

and hence there is a constant c_{2} such that

exp (−i\frac{(t − s) |p_{t} − p_{s}|^{2}}{4 + (t − s)^{2}} ) = exp(i(s − t + sgn(s − t)θγ^{2})) + O(exp(c_{2}γ^{2}⟨t − s⟩^{−1}) − 1).

Therefore, for some constant c_{3},

K(t, s) = \frac{−16 exp(−2)(t − s)^{2}}{4π(−2 + i(t − s))^{3}} exp(i(s − t)) exp(i sgn(s − t)θγ^{2})
+ O(γ^{4}⟨t − s⟩^{−3} + ⟨t − s⟩^{−2}(exp(c_{3}γ^{2}⟨t − s⟩^{−1}) − 1)).

Hence

E(K(t, s)) = \frac{−16 exp(−2)(t − s)^{2}}{4π(2 − i(t − s))^{3}} exp(i(s − t))(a_{1} + ia_{2} sgn(s − t)) + O(⟨t − s⟩^{−3}),

where a_{1} + ia_{2} = E(exp(iθγ^{2})). By considering the examples α_{t} = exp(it)/\sqrt{N} if |t| ≤ N, and 0 otherwise, and letting N → ∞, it may be readily seen that E(K(t, s)) is not the kernel of a bounded map from L_{2}(R) → L_{2}(R), and the desired counterexample has been obtained.

We will now present a second proof of the same result. This proof goes via the Fourier transform. Although this proof is less intuitive, it also has less technical difficulties. To recap, it is sufficient to find a counterexample to the following statement: there is a constant c such that if ∫_{−∞}^{∞} |α_{t}|^{2} dt ≤ 1, and if g ∈ H_{1}(R^{2}), then for any path p_{t} ∈ R^{2}, we have that

∫_{−∞}^{∞} ∫_{−∞}^{∞} K(s, t)α_{t}ᾱ_{s} ds dt ≤ c,

where

K(s, t) = ∫_{R^{2}}(A_{t−s}g)(x − p_{t})\overline{g(x − p_{s})} dx ds dt
= ∫_{R^{2}} exp(i(t − s) |ζ|^{2} + i(p_{s} − p_{t}) · ζ) |ĝ(ζ)|^{2} dζ.

That is, we are asking whether K(s, t) is the kernel of a bounded operator from L_{2}(R) to L_{2}(R).

We take p_{t} = (t, b_{t}), where b_{t} is Brownian motion. It is sufficient to show that E(K(s, t)) is not the kernel of a bounded operator on L_{2}(R^{2}). Note that if γ is a gaussian random variable with mean 1 and standard deviation 0, then E(exp(iγ)) = exp(−t^{2}/2). Hence E(K(s, t)) = L(t − s), where

L(t) = ∫_{R^{2}} exp(it |ζ|^{2} − itζ_{1} − |t| ζ_{2}^{2}/2) |ĝ(ζ)|^{2} dζ.

But L(t − s) fails to be the kernel of a bounded operator on L_{2}(R) if and only if L̂(ω) fails to be bounded almost everywhere. Furthermore

L̂(ω) = ∫_{R^{2}} k(ω, ζ) |g(ζ)|^{2} dζ,

where k(·, ζ) = k_{1}(·, ζ) ∗ k_{2}(·, ζ) with

k_{1}(ω, ζ) = δ(ω + |ζ|^{2} − ζ_{1})

and

k_{2}(ω, ζ) = \frac{4ζ_{2}^{2}}{4ω^{2} + ζ_{2}^{4}} ,

that is,

k(ω, ζ) = \frac{4ζ_{2}^{2}}{4(ω + |ζ|^{2} − ζ_{1})^{2} + ζ_{2}^{4}} .

Clearly k enjoys enough continuity properties so that it is sufficient to show that L̂(ω) is unbounded for ω = 1/4. So

k(1/4, ζ) = \frac{4ζ_{2}^{2}}{4((ζ_{1} − 1/2)^{2} + ζ_{2}^{2})^{2} + ζ_{2}^{4}} ≥ \frac{ζ_{2}^{2}}{2 |ζ − (1/2, 0)|^{4}} .

It is clear that k(1/4, ζ) has an L_{1}(R^{2}) singularity at (1/2, 0), and hence taking ĝ(ζ) = |ζ|^{2} exp(− |ζ|^{2}), we see that L̂(1/4) is unbounded.

3. Solutions of the wave equation This section is devoted to the following result.

Theorem 3. Assertion (B_{3,BMO,2}) is not true.

We will show that assertion (B_{3,H_{1},2}^{∗}) is not true. The methods will be essentially the same as in the previous proofs, but the details will be more difficult, and so we will break it into steps.

To start, let us define an operator on a dense subspace of L_{2}(R^{3}) for each t ∈ R given by

Ĉ_{t}f(ζ) = \frac{cos(t |ζ|)}{|ζ|^{2}} f̂(ζ).

Let us also set

K_{t}(x) = \frac{1}{4π |x|}I_{|x|≥|t|} (x ∈ R^{3}).

Lemma 4. If f ∈ L_{1}(R^{3}) ∩ L_{2}(R^{3}), and t ≠ 0, then

C_{t}f(x) = ∫_{R^{3}} K_{t}(x − y)f(y) dy,

and the operator norm of C_{t} from L_{1}(R^{3}) to L_{1}(R^{3}) is (4π |t|)^{−1}.

To show this when f is C^{∞} with compact support, it is sufficient to show that K̂_{t} = cos(t |ζ|)/ |ζ|^{2} (as tempered distributions). Then by Young’s convolution formula, as an operator from L_{1}(R^{3}) to L_{1}(R^{3}), the operator norm of C_{t} is given by

∥C_{t}∥ = ∥K_{t}∥_{∞} = \frac{1}{4π |t|}.

It is clear that K̂_{t}(ζ) depends only upon t and |ζ|. So without loss of generality, we may suppose that ζ = (ζ_{1}, 0, 0), where ζ_{1} = |ζ|. In performing the following integral, we will use the following change of variables: x_{1} = u, x_{2} = \sqrt{v^{2} − u^{2}} cos(θ), and x_{3} = \sqrt{v^{2} − u^{2}} sin(θ). Thus the Jacobian ∂(x_{1}, x_{2}, x_{3})/∂(u, v, θ) = v = |x|. Then

K̂_{t}(ζ) = ∫_{|x|≥|t|} \frac{exp(−ix · ζ)}{4π |x|} dx

= \frac{1}{4π} ∫_{v=|t|}^{∞} ∫_{u=−v}^{v} ∫_{0}^{2π} cos(u |ζ|) dθ dv du

= \frac{cos(t |ζ|)}{|ζ|^{2}}.

In the last line we have used the assertion that lim_{R→∞} cos(R |ζ|) = 0, which is true in the space of tempered distributions.

Corollary 5. If f ∈ L_{2}(R^{3}) is bounded with compact support, and s, t ∈ R, then

B_{s}B_{t}f(x) = \frac{C_{s+t}f(x) − C_{s−t}f(x)}{2} = ∫_{R^{3}} \frac{K_{s+t}(x − y) − K_{s−t}(x − y)}{2}f(y) dy.

This follows because sin(t |ζ|) sin(s |ζ|) = \frac{1}{2}(cos((s + t) |ζ|) − cos((s − t) |ζ|)).

Lemma 6. There is a universal constant c such that if t ↦ f_{t} is in L_{2}(R, L_{1}(R^{3})), then

∫_{0}^{∞} ∫_{0}^{∞} ∫_{R^{3}} ∫_{R^{3}} K_{s+t}(x − y)f_{t}(x)f_{s}(y) dy dx dt ds ≤ c ∫_{0}^{∞} ∥f_{t}∥_{1}^{2} dt.

The proof of this result depends upon the boundedness of the Hardy operators defined on L_{2}([0, ∞)):

Hα(t) = \frac{1}{t} ∫_{0}^{t} α(s) ds
H^{∗}α(t) = ∫_{t}^{∞} \frac{α(s)}{s} ds.

If α ∈ L_{2}([0, ∞)), then both Hα and H^{∗}α are in L_{2}([0, ∞)), with ∥Hα∥_{2} , ∥H^{∗}α∥_{2} ≤ 2 ∥α∥_{2}. The result for H may be found in [Ha], and the result for H^{∗} follows because H^{∗} is the adjoint operator to H.

Next, by Lemma 4, if s, t > 0

∫_{R^{3}} ∫_{R^{3}} K_{s+t}(x − y)f_{t}(x)f_{s}(y) dy dx ≤ \frac{1}{4π(s + t)} ∥f_{t}∥_{1} ∥f_{s}∥_{1} .

Thus, setting α(t) = ∥f_{t}∥_{1}, we see that

∫_{0}^{∞} ∫_{0}^{∞} ∫_{R^{3}} ∫_{R^{3}} K_{s+t}(x − y)f_{t}(x)f_{s}(y) dy dx dt ds

≤ \frac{1}{4π} ∫_{0}^{∞}(Hα(t) + H^{∗}α(t))α(t) dt

≤ \frac{1}{π} ∥α∥_{2}^{2} .

Now we will consider the following assertion. (C): There is a universal constant c such that if t ↦ f_{t} is in L_{2}(R, H_{1}(R^{3})), then

∫_{−∞}^{∞} ∫_{−∞}^{∞} ∫_{R^{3}} ∫_{R^{3}} K_{s−t}(x − y)f_{t}(x)f_{s}(y) dy dx dt ds ≤ c ∫_{−∞}^{∞} ∥f_{t}∥_{H_{1}}^{2} dt.

Let us first demonstrate that the failure of assertion (C) implies the failure of assertion (B_{3,H_{1},2}). Let us suppose that we have a sequence t ↦ f_{t} in L_{2}(R, H_{1}(R^{3})) such that

I = ∫_{−∞}^{∞} ∫_{−∞}^{∞} ∫_{R^{3}} ∫_{R^{3}} K_{s−t}(x − y)f_{t}(x)f_{s}(y) dy dx dt ds

is unbounded. We may split the integral up into four pieces:

I = I_{1} + I_{2} + I_{3} + I_{4}

= ∫_{0}^{∞} ∫_{0}^{∞} … dt ds + ∫_{−∞}^{0} ∫_{0}^{∞} … dt ds + ∫_{0}^{∞} ∫_{−∞}^{0} … dt ds + ∫_{−∞}^{0} ∫_{−∞}^{0} … dt ds.

By Lemma 6, we know that I_{2} and I_{3} are bounded. Therefore one of I_{1} or I_{4} is unbounded, and without loss of generality we may suppose that I_{1} is unbounded. So without loss of generality, we may suppose that f_{t} = 0 if t < 0.

Now, multiplying out, and applying Corollary 5, we see that

∥∫_{−∞}^{∞} B_{t}f_{t} dt∥_{2}^{2} = ∥∫_{0}^{∞} B_{t}f_{t} dt∥_{2}^{2}

= ∫_{0}^{∞} ∫_{0}^{∞} ∫_{R^{3}} ∫_{R^{3}} \frac{K_{s+t}(x − y) − K_{s−t}(x − y)}{2}f_{t}(x)f_{s}(y) dy dx dt ds.

Again, by Lemma 6, the last integral differs from I_{1}/2 by a bounded amount, and we have produced the desired counterexample.

All that remains to be shown is the following.

Lemma 7. Assertion (C) is false.

Our counterexample will be f_{t}(x) = α_{t}g(x − p_{t}), where ∫_{0}^{∞} |α_{t}|^{2} dt = 1, and p_{t} = (t, b_{t}, b′_{t}). The function g ∈ H_{1}(R^{3}) will be selected later. Here b_{t} and b′_{t} are two independent Brownian motions. We will show that

J = E (∫_{−∞}^{∞} ∫_{−∞}^{∞} ∫_{R^{3}} ∫_{R^{3}} K_{s−t}(x − y)α_{t}ᾱ_{s}g(x − p_{t})\overline{g(x − p_{s})} dy dx dt ds)

cannot be universally bounded. This quantity is more easily computed using the Fourier transform:

J = E (∫_{−∞}^{∞} ∫_{−∞}^{∞} α_{t}ᾱ_{s} ∫_{R^{3}} \frac{cos((t − s) |ζ|)}{|ζ|^{2}} exp(i(p_{s} − p_{t}) · ζ) |ĝ(ζ)|^{2} dζ dt ds) .

Notice that

E(exp(i(p_{s} − p_{t})) = exp(i(s − t)ζ_{1})E(exp(i(b_{s} − b_{t})ζ_{2}) + i(b′_{s} − b′_{t})ζ_{3})
= exp(i(s − t)ζ_{1}) exp(− |s − t| (ζ_{2}^{2} + ζ_{3}^{2})/2).

Hence

J = ∫_{−∞}^{∞} ∫_{−∞}^{∞} L(s − t)α_{t}ᾱ_{s} dt ds,

where

L(t) = ∫_{R^{3}} \frac{cos(t |ζ|)}{|ζ|^{2}} exp(itζ_{1}) exp(− |t| (ζ_{2}^{2} + ζ_{3}^{2})/2) |ĝ(ζ)|^{2} dζ.

Saying that J is bounded for all α_{t} in L_{2}(R) is equivalent to saying that convolving with L gives a bounded operator on L_{2}(R). But this is equivalent to L̂ being in L_{∞}(R). Now

L̂(ω) = ∫_{R^{3}} k(ω, ζ) |ĝ(ζ)|^{2} / |ζ|^{2} dζ,

where k(·, ζ) = h_{1}(·, ζ) ∗ h_{2}(·, ζ) ∗ h_{3}(·, ζ) with

h_{1}(ω, ζ) = \frac{δ(ω + |ζ|) + δ(ω − |ζ|)}{2}
h_{2}(ω, ζ) = δ(ω − ζ_{1})
h_{3}(ω, ζ) = \frac{4(ζ_{2}^{2} + ζ_{3}^{2})}{4ω^{2} + (ζ_{2}^{2} + ζ_{3}^{2})^{2}} .

Thus

k(ω, ζ) = \frac{k_{1}(ω, ζ) + k_{2}(ω, ζ)}{2},

where

k_{1}(ω, ζ) = \frac{4(ζ_{2}^{2} + ζ_{3}^{2})}{4(ω − ζ_{1} + |ζ|)^{2} + (ζ_{2}^{2} + ζ_{3}^{2})^{2}} ,

and

k_{2}(ω, ζ) = \frac{4(ζ_{2}^{2} + ζ_{3}^{2})}{4(ω − ζ_{1} − |ζ|)^{2} + (ζ_{2}^{2} + ζ_{3}^{2})^{2}} .

Since all the expressions involved are positive, we will have found a counterexample if we can show that

∫_{U} k_{1}(0, ζ) |ĝ(ζ)|^{2} / |ζ^{2}| dζ = ∞,

where U is any subset of R^{3}. We will take

U = \{ζ : \sqrt{ζ_{2}^{2} + ζ_{3}^{2}} ≤ ζ_{1} ≤ 1\},

and

g(x) = \{ \begin{matrix}1 & if 0 < x_{1} ≤ 1 and −1 ≤ x_{2}, x_{3} ≤ 1 \\ −1 & if −1 ≤ x_{1} < 0 and −1 ≤ x_{2}, x_{3} ≤ 1 \\ 0 & otherwise, \end{matrix}

so that

ĝ(ζ) = \frac{8(1 − cos(ζ_{1})) sin(ζ_{2}) sin(ζ_{3})}{ζ_{1}ζ_{2}ζ_{3}}.

In this case, we see that |ĝ(ζ)|^{2} / |ζ|^{2} is bounded from below by a positive number on U. Therefore, we need to show that

∫_{U} k_{1}(0, ζ) dζ = ∞.

To do this, let us compute the integral using cylindrical coordinates, that is, we will write z = ζ_{1}, and r = \sqrt{ζ_{2}^{2} + ζ_{3}^{2}}. Then we see that the last integral is

∫_{z=0}^{1} ∫_{r=0}^{z} \frac{4r^{2}}{4(\sqrt{z^{2} + r^{2}} − z)^{2} + r^{4}} 2πr dr dz.

But in U,

\sqrt{z^{2} + r^{2}} − z ≤ r^{2}/z,

and hence

\frac{4r^{2}}{4(\sqrt{z^{2} + r^{2}} − z)^{2} + r^{4}} ≥ \frac{4z^{2}}{r^{2}(4 + z^{2})}.

Hence the integral in question becomes bounded below by

∫_{z=0}^{1} ∫_{r=0}^{z} \frac{8πz^{2}}{r(4 + z^{2})} dr dz,

and this is easily seen to be infinite.

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S.J. Montgomery-Smith Mathematics Department University of Missouri Columbia, MO 65211 E-mail: stephen@math.missouri.edu