SOLUTIONS TO HW 7

TERENCE TAO

1. Exercise 2.28 (Pseudoconformal transformation)

We can write v as

v(t, x) = (2π)^{d/2}K_{t}(x)\overline{u(\frac{1}{t}, \frac{x}{t})}

where K_{t} is the fundamental solution (note that this is compatible with the fact that K_{t} and 1 both solve the Schrödinger equation for t ≠ 0). We can then use the product rule to obtain

(i∂_{t} + \frac{1}{2}Δ)v(t, x) = (2π)^{d/2}[(i∂_{t} + \frac{1}{2}Δ)K_{t}\overline{u(\frac{1}{t}, \frac{x}{t})}

+ K_{t}(i∂_{t} + \frac{1}{2}Δ)[\overline{u(\frac{1}{t}, \frac{x}{t})}]

+ ∇K_{t} · ∇[\overline{u(\frac{1}{t}, \frac{x}{t})}].

Since K_{t} is the fundamental solution, we have (i∂_{t} + \frac{1}{2}Δ)K_{t} = 0. Also we observe that ∇K_{t} = \frac{ix}{t}K_{t}. So it suffices to show that

(i∂_{t} + \frac{1}{2}Δ)[\overline{u(\frac{1}{t}, \frac{x}{t})}] + \frac{ix}{t}∇[\overline{u(\frac{1}{t}, \frac{x}{t})} = \frac{1}{t^{2}} \overline{iu_{t} + \frac{1}{2}Δu}(\frac{1}{t}, \frac{x}{t}).

But this follows from the chain rule.

Now let us investigate what happens as t → 0. From the fundamental solution we have

u(t, x) = \frac{1}{(2πit)^{d/2}} ∫_{R_{d}} e^{i|x−y|^{2}/2t}u_{0}(y) dy

and hence on substitution

v(t, x) = \frac{1}{(2π)^{d/2}} e^{i|x|^{2}/2t}\overline{∫_{R_{d}} e^{it|x/t−y|^{2}/2}u_{0}(y) dy}

which simplifies to

v(t, x) = \frac{1}{(2π)^{d/2}} ∫_{R_{d}} e^{ix·y}e^{−it|y|^{2}/2}\overline{u_{0}}(y) dy

and so it is clear (since u_{0} is Schwartz and dominated convergence) that

lim_{t→0} v(t, x) = \frac{1}{(2π)^{d/2}} ∫_{R_{d}} e^{ix·y}\overline{u_{0}}(y) dy = \frac{1}{(2π)^{d/2}} \overline{û_{0}}(x).

To show that v is in fact smooth at t = 0 and solves the equation, consider the function v_{+}, defined by (say) v_{+}(t) := e^{i(t−1)Δ/2}v(1) (i.e. we solve the Schrödinger equation starting from time t = 1. Since v(1) is Schwartz (by inspection), we see that v_{+} is also Schwartz. From uniqueness for the Schrödinger equation in the Schwartz category, we see that v(t) = v_{+}(t) for all t > 0, and hence by continuity v(0) = v_{+}(0). In particular this lets us conclude that v(t) = v_{+}(t) = e^{itΔ/2}v_{+}(0) = e^{itΔ/2}v(0) for t ≥ 0. A similar argument gives v(t) = e^{itΔ/2}v(0) for t ≤ 0, and hence v solves the Schrödinger equation for all t.

2. Exercise 3.15 (H^{k,k} is a Banach algebra)

We may normalise u and v to have H_{x}^{k,k}(R^{d}) norm one, thus

∥⟨x⟩^{j}u∥_{H_{x}^{k−j}(R^{d})}, ∥⟨x⟩^{j}v∥_{H_{x}^{k−j}(R^{d})} ≤ 1 for 0 ≤ j ≤ k.  (1)

Now our task is to show that

∥⟨x⟩^{j}uv∥_{H_{x}^{k−j}(R^{d})} ≲_{k,d} 1

for all 0 ≤ j ≤ k (since we can then sum in j to obtain the result. If we write w := ⟨x⟩^{j}u, it would thus be enough to show that

∥wv∥_{H_{x}^{k−j}(R^{d})} ≲_{k,d} ∥w∥_{H_{x}^{k−j}}(R^{d})∥v∥_{H_{x}^{k}(R^{d})}.

Now if j = 0, this follows from Lemma A.8 (since H^{k} controls L^{∞}); if j = k, this follows from Hölder’s inequality (again using that H^{k} controls L^{∞}). The intermediate cases then follow by complex interpolation.

OK, so that was perhaps a bit too slick. Here is a more pedestrian approach. Expanding out the definitions, we have to show that

∥∇^{m}(⟨x⟩^{j}uv)∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1

whenever 0 ≤ j ≤ k and 0 ≤ m ≤ k − j. Using the Leibnitz rule and the triangle inequality (and noting that |∇^{r}⟨x⟩^{j}| ≲_{r,j,d} ⟨x⟩^{j−r}), it suffices to show that

∥⟨x⟩^{j}|∇^{m_{1}}u||∇^{m_{2}}v|∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1

whenever 0 ≤ j, m_{1}, m_{2}, and j + m_{1} + m_{2} ≤ k.

To do this it is convenient to induct on j + m_{1} + m_{2}. When j + m_{1} + m_{2} = 0 we have to show that

∥uv∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1

but this is easy, as we can (say) place u in L_{x}^{2}, v in L_{x}^{∞} and use Sobolev (we know that v is bounded in H_{x}^{k}(R^{d}), and hence in L_{x}^{∞} since k > d/2). Now suppose that j + m_{1} + m_{2} > 0 and the claim has already been proven for smaller values of j + m_{1} + m_{2}.

Without loss of generality we can take m_{1} ≤ m_{2}. We can use the Leibnitz rule to write

⟨x⟩^{j}∇^{m_{1}}u = ∇^{m_{1}}(⟨x⟩^{j}u) + l.o.t.

where the lower order terms will be treatable by the induction hypothesis. So it will suffice to show that

∥|∇^{m_{1}}(⟨x⟩^{j}u)||∇^{m_{2}}v|∥_{L_{x}^{2}(R^{d})} ≲_{k,d} 1.

We split the left-hand side using Hölder as

∥∇^{m_{1}}(⟨x⟩^{j}u)∥_{L_{x}^{p}(R^{d})}∥∇^{m_{2}}v∥_{L_{x}^{q}(R^{d})}

where p, q are exponents with 1/p+1/q = 1/2 that we shall choose presently. Since ⟨x⟩^{j}u is known to be in H^{k−j}, Sobolev embedding allows us to bound the L^{p} norm so long as \frac{d}{2} − \frac{d}{p} ≤ k − j − m_{1}; since v is known to be in H^{k}, we can similarly bound the L^{q} norm so long as \frac{d}{2} − \frac{d}{q} ≤ k − m_{2}. Since k − j − m_{1} and k − m_{2} are non-negative and sum to at least d, we see that we can choose p and q that simultaneously satisfy these bounds.

UCLA Department of Mathematics, Los Angeles, CA 90095-1596.

E-mail address: tao@math.ucla.edu