SOLUTIONS TO HW 1

TERENCE TAO

1. Exercise A.8 (Relationship between Sobolev and isoperimetric inequalities)

Incidentally, one needs to implicitly assume d > 1 in order for things to make sense in this question.

To establish the inequality

λ^{d/(d−1)}|{x : |f(x)| ≥ λ}| ≲_{d} ∥∇f∥_{L^{1}(R^{d})}^{d/(d−1)}  (1)

there are several approaches. One way, as indicated in the hint, is to first establish the pointwise inequality

|f(x)| ≲_{d} ∫_{R_{d}} \frac{|∇f(y)|}{|x − y|^{d−1}} dy.  (2)

This in turn can be established in a number of ways. One is by starting with the Newton potential formula

f(x) = c_{d} ∫_{R_{d}} \frac{Δf(y)}{|x − y|^{d−2}} dy

for some c_{d} > 0 and integrating by parts (this only works as stated for d > 2; the case d = 2 needs to be handled slightly differently). Another is by observing from the fundamental theorem of calculus that

|f(x)| = | ∫_{0}^{∞} ∂_{r}f(x + rω) dr|
≤ ∫_{0}^{∞} |∇f(x + rω)| dr

for every direction ω ∈ S^{d−1}; averaging this inequality over all ω and switching from polar coordinates back to Cartesian, one obtains

|f(x)| ≲_{d} ∫_{S^{d−1}} ∫_{0}^{∞} |∇f(x + rω)| drdω

= ∫_{R_{d}} \frac{|∇f(x + y)|}{|y|^{d−1}} dy

= ∫_{R_{d}} \frac{|∇f(y)|}{|x − y|^{d−1}} dy

as desired. (Note that as f is Schwartz, all expressions here are absolutely integrable and there is no difficulty justifying the computations.)

Once one has (2), one can prove (1) either by appeal to the endpoint Hardy-Littlewood-Sobolev theorem of fractional integration (as for instance in Stein’s “singular integrals” - note that the non-endpoint theorem in the notes is not quite applicable at L^{1}, and indeed convolution by \frac{1}{|x|^{d−1}} does NOT map L^{1} to L^{d/(d−1)}, only to weak L^{d/(d−1)}) by the following duality argument. Let E := {x : |f(x)| ≥ λ}, thus our task is to show that

λ^{d/(d−1)}|E| ≲_{d} ∥∇f∥_{L^{1}(R^{d})}^{d−1}.  (3)

(One can use scaling and homogeneity symmetries to normalise two of the three quantities λ, |E|, ∥∇f∥_{L^{1}(R^{d})} to equal one, but we will refrain from doing so here.) Since f is Schwartz, E is bounded and measurable. Now consider the quantity

∫_{R_{d}} 1_{E}(x)|f(x)| dx.  (4)

On the one hand, since |f(x)| ≥ λ on E, this integral is at least λ|E|. On the other hand, we have by (2) and Fubini

∫_{R_{d}} 1_{E}(x)|f(x)| dx ≲_{d} ∫_{R_{d}} 1_{E}(x) ∫_{R_{d}} \frac{|∇f(y)|}{|x − y|^{d−1}} dydx

= ∫_{R_{d}} |∇f(y)|(∫_{E} \frac{1}{|x − y|^{d−1}} dx) dy

≤ ∥∇f∥_{L^{1}(R^{d})} sup_{y∈R^{d}}(∫_{E} \frac{1}{|x − y|^{d−1}} dx).

Now we observe the estimate

∫_{E} \frac{dx}{|x − y|^{d−1}} ≲_{d} |E|^{1/d}

for any fixed y ∈ R^{d}. Intuitively, the point here is that for a fixed amount of measure |E|, the way that one would choose E to maximise the integral on the left is to make E a ball centred at y of radius ∼_{d} |E|^{1/d}, at which point the estimate becomes clear. To make the intuition more precise, what we can do is pick a cutoff radius r > 0 and split

∫_{E} \frac{dx}{|x − y|^{d−1}} = ∫_{E:|x−y|<r} \frac{dx}{|x − y|^{d−1}} + ∫_{E:|x−y|≥r} \frac{dx}{|x − y|^{d−1}} .

For the first term, we throw away E and estimate by ∫_{|x−y|<r} \frac{dx}{|x−y|^{d−1}} ∼_{d} r; for the second term, we estimate \frac{1}{|x−y|^{d−1}} by \frac{1}{r^{d−1}} and estimate by |E|/r^{d−1}. Thus we have

∫_{E} \frac{dx}{|x − y|^{d−1}} ≲_{d} r + |E|/r^{d−1};

we optimise in r (by setting r := |E|^{1/d}, which is consistent with the intuition) to obtain the result. Putting together the lower and upper bounds for (4) we obtain

λ|E| ≲_{d} ∥∇f∥_{L^{1}(R^{d})}|E|^{1/d}

which after some algebra gives the result.

One can also prove (3) by Littlewood-Paley theory. Let N be a frequency to be chosen later, and write

f = P_{<N}f + P_{≥N}f.

For the low frequencies, we have Bernstein’s inequality

∥P_{<N}f∥_{L^{∞}(R^{d})} ≤ ∑_{M<N} ∥P_{M}f∥_{L^{∞}(R^{d})}

≲_{d} ∑_{M<N} M^{d}∥P_{M}f∥_{L^{1}(R^{d})}

≲_{d} ∑_{M<N} M^{d−1}∥∇f∥_{L^{1}(R^{d})}

≲_{d} N^{d−1}∥∇f∥_{L^{1}(R^{d})}.

Thus if we choose

N := c_{d}(λ/∥∇f∥_{L^{1}(R^{d})})^{1/(d−1)}

for some sufficiently small c_{d} > 0, we have

|P_{<N}f(x)| ≤ λ/2 for all x ∈ R^{d}

and hence

{x ∈ R^{d} : |f(x)| ≥ λ} ⊆ {x ∈ R^{d} : |P_{≥N}f(x)| ≥ λ/2}.

Now we use Chebyshev’s inequality:

|\{x ∈ R^{d} : |P_{≥N}f(x)| ≥ λ/2\}| ≤ \frac{2}{λ}∥P_{≥N}f∥_{L^{1}(R^{d})}

≲ \frac{1}{λ} ∑_{M≥N} ∥P_{M}f∥_{L^{1}(R^{d})}

≲_{d} \frac{1}{λ} ∑_{M≥N} M^{−1}∥∇f∥_{L^{1}(R^{d})}

≲ \frac{1}{λ}N^{−1}∥∇f∥_{L^{1}(R^{d})}

≲_{d} λ^{−d/(d−1)}∥∇f∥_{L^{1}(R^{d})}^{d/(d−1)}

as desired, where we used the formula for N.

A rather different (and more elementary approach) to the inequality is to use Chebyshev immediately and prove

∥f∥_{L^{d/(d−1)}(R^{d})} ≲_{d} ∥∇f∥_{L^{1}(R^{d})}.  (5)

Here one cannot use (2) because the convolution with 1/|x|^{d−1} does not map L^{1} to L^{d/(d−1)} (basically because 1/|x|^{d−1} itself is not bounded in L^{d/(d−1)}). However one can argue by the following much more elementary argument. We induct on d. For d = 1 the estimate

∥f∥_{L^{∞}(R)} ≲ ∥∇f∥_{L^{1}(R)}

is clear from the fundamental theorem of calculus. Now suppose that d > 1 and the claim has already been shown for d − 1. We parameterise R^{d} as (x, t) where x ∈ R^{d−1} and t ∈ R. We use the fundamental theorem of calculus to estimate

|f(x, t)| ≤ ∫R |∇f(x, t′)| dt′

and hence by Fubini’s theorem we see that for any fixed t we have

∥f(x, t)∥_{L_{x}^{1}(R^{d−1})} ≤ ∥∇f∥_{L^{1}(R^{d})}.

On the other hand, from induction hypothesis we see that for any fixed t we have

∥f(x, t)∥_{L_{x}^{(d−1)/(d−2)}(R^{d−1})} ≲_{d} ∥∇f(x, t)∥_{L_{x}^{1}(R^{d−1})}

and hence by Hölder’s inequality

∥f(x, t)∥_{L_{x}^{d/(d−1)}(R^{d−1})} ≲_{d} ∥∇f(x, t)∥_{L_{x}^{1}(R^{d−1})}^{(d−1)/d}∥∇f∥_{L^{1}(R^{d})}^{1/d}.

Raising both sides to the power d/(d − 1) and integrating we obtain (5) as desired.

Now we want to apply this inequality to study the isoperimetric inequality for a smooth bounded domain Ω. Formally, the way one should proceed is as follows. If we set f = 1_{Ω}, then formally |∇f| is surface measure on ∂Ω and hence ∥∇f∥_{L^{1}} is equal to the surface area of Ω. Meanwhile, |{x ∈ R^{d} : |f(x)| ≥ 1}| is just the volume of Ω, and the claim follows immediately.

There is an issue because the Sobolev inequality was only established for Schwartz functions, and 1_{Ω} is not smooth. So one has to regularise 1_{Ω}. There are a number of ways to do this; some of you used the tubular neighbourhood theorem or the coarea formula, which certainly work here. A more pedestrian approach is to take a standard approximation to the identity ψ_{ε}(x) := ε^{−d}ψ(x/ε), where ψ is a nonnegative bump function supported on B(0, 1) with total mass 1, and consider the regularisations f_{ε} := 1_{Ω} ∗ ψ_{ε}. This is Schwartz (indeed, it is smooth and compactly supported) and so we have

|{x ∈ R^{d} : f_{ε}(x) ≥ 1}| ≲_{d} ∥∇f_{ε}∥_{L^{1}(R^{d})}^{d/(d−1)}.

Now if B(x, ε) lies completely inside Ω we see that f_{ε}(x) = 1, so we see that

lim inf_{ε→0} |{x ∈ R^{d} : f_{ε}(x) ≥ 1}| ≥ |Ω|.

(Indeed, the limit exists and is equal to |Ω|, but we do not need this). To conclude the argument we just need to show that

lim inf_{ε→0} ∥∇f_{ε}∥_{L^{1}(R^{d})} ≲_{d} |∂Ω|.  (6)

One way to do this is as follows. Observe that

∇f_{ε} = ∇(1_{Ω} ∗ ψ_{ε}) = 1_{Ω} ∗ ∇ψ_{ε}

and hence by Young’s inequality

∥∇f_{ε}∥_{L^{∞}(R^{d})} ≤ ∥1_{Ω}∥_{L^{∞}(R^{d})}∥∇ψ_{ε}∥_{L^{1}(R^{d})} = ∥∇ψ_{ε}∥_{L^{1}(R^{d})}.

Now direct computation shows that ∇ψ_{ε} has size O_{d}(ε^{−d−1}) on a ball of radius ε, and vanishes elsewhere, thus

∥∇f_{ε}∥_{L^{∞}(R^{d})} ≤ ∥∇ψ_{ε}∥_{L^{1}(R^{d})} ≲_{d} ε^{−1}.

Further direct computation shows that ∇f_{ε} vanishes except in the ε-neighbourhood ∂Ω_{ε} of ∂Ω, thus

∥∇f_{ε}∥_{L^{1}(R^{d})} ≲ \frac{1}{ε}|∂Ω_{ε}|.

As ε → 0, the right-hand side is bounded by a constant multiple of the surface area of ∂Ω, and the claim follows.

There are other regularisations that are available. One approach uses the metric function d(x, Ω), which is smooth in a small annular neighbourhood of Ω; one can take f to be some suitable function of d(x, Ω), such as f_{ε}(x) = η(d(x, Ω)/ε) where η(t) is a smooth function which equals 1 for t < 1 and equals 0 for t ≥ 2. The advantage of working with d is that it is automatically Lipschitz (by the triangle inequality) and so one gets good bounds on f_{ε} (actually, much the same bounds as before). Another approach is to use a coordinatisation given by the tubular neighbourhood theorem.

2. Exercise A.13 (Fractional chain rule)

We first establish the bounds for a single Littlewood-Paley piece:

∥P_{N}F (f)∥_{W^{s,q}(R^{d})} ≲_{d,p,q,r,s} ∥f∥_{W^{s,r}(R^{d})}^{p}  (7)

and

∥P_{N}(F (f) − F (g))∥_{W^{s,q}(R^{d})} ≲_{d,p,q,r,s} (∥f∥_{W^{s,r}(R^{d})} + ∥g∥_{W^{s,r}(R^{d})})^{p−1}∥f − g∥_{W^{s,r}(R^{d})}. (8)

We begin with the first bound (7). One could use homogeneity (rescaling F appropriately) to normalise ∥f∥_{W^{s,r}(R^{d})} = 1 but we will not do so here. To simplify the notation we allow all implied constants in the ≲ notation to depend on d, p, q, r, s.

The case s = 0 follows immediately from Hölder’s inequality, so we take s > 0.

It will be convenient to introduce the exponents t defined by

\frac{d}{t} = \frac{d}{r} − s;

from the hypotheses on q and r we see that r < t < ∞. From Sobolev embedding we have

∥f∥_{L^{t}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}.  (9)

Also from the scaling condition we see that

\frac{1}{q} = \frac{p − 1}{t} + \frac{1}{r}.  (10)

The significance of the exponent t can already be seen by the heuristic proof of the fractional chain rule using Principle A.6.

The case N ≤ 1 is rather easy; in that case we estimate

∥P_{N}F (f)∥_{W^{s,q}(R^{d})} ≲ ∥F (f)∥_{L^{q}(R^{d})}
≲ ∥|f|^{p}∥_{L^{q}(R^{d})}
≲ ∥f∥_{L^{qp}(R^{d})}^{p}
≲ ∥f∥_{W^{s,r}(R^{d})}^{p}

where the last bound will follow from Sobolev embedding. For the case N > 1, we write

∥P_{N}F (f)∥_{W^{s,q}(R^{d})} ≲ N^{s}∥P_{N}F (f)∥_{L^{q}(R^{d})}.

Now we split f = P_{<N}f + P_{≥N}f. Taylor expansion is a bit too expensive in F (we only are assuming C^{1} bounds rather than C^{2}), so we turn to either the mean-value theorem or the fundamental theorem of calculus. Let’s use the latter (which is a bit more precise, and also works in the complex case in contrast to the mean-value theorem):

F (f) = F (P_{<N}f)+∫_{0}^{1} \frac{d}{dθ} F (P_{<N}f+θP_{≥N}f) dθ = F (P_{<N}f)+∫_{0}^{1} F′(P_{<N}f+θP_{≥N}f)P_{≥N}f dθ.

If f is complex-valued one has to replace F′ with F_{z} and F\overline{z}, but we omit this to keep the exposition simple. Applying the triangle and Minkowski we reduce to showing

∥P_{N}F (P_{<N}f)∥_{L^{q}(R^{d})} ≲ N^{−s}∥f∥_{W^{s,r}(R^{d})}^{p}

and

∥P_{N}[F′(P_{<N}f + θP_{≥N}f)P_{≥N}f]∥_{L^{q}(R^{d})} ≲ N^{−s}∥f∥_{W^{s,r}(R^{d})}^{p}

for all 0 ≤ θ ≤ 1.

For the first bound, we exploit the smoothness of P_{<N}f (and hence of F (P_{<N}f)) by differentiation, as well as (10) and (9):

∥P_{N}F (P_{<N}f)∥_{L^{q}(R^{d})} ≲ N^{−1}∥∇F (P_{<N}f)∥_{L^{q}(R^{d})}

= N^{−1}∥F′(P_{<N}f)∇P_{<N}f∥_{L^{q}(R^{d})}

≲ N^{−1}∥|P_{<N}f|^{p−1}|∇P_{<N}f|∥_{L^{q}(R^{d})}

≲ N^{−1}∥P_{<N}f∥_{L^{t}(R^{d})}^{p−1}∥∇P_{<N}f∥_{L^{r}(R^{d})}

≲ N^{−1}∥f∥_{L^{t}(R^{d})}^{p−1} ∑_{M<N} ∥∇P_{M}f∥_{L^{r}(R^{d})}

≲ N^{−1}∥f∥_{W^{s,r}(R^{d})}^{p−1} ∑_{M<N} M^{1−s}∥f∥_{W^{s,r}(R^{d})}

≲ N^{−s}∥f∥_{W^{s,r}(R^{d})}^{p} as desired. For the second bound, we use crude pointwise estimates:

∥P_{N}[F′(P_{<N}f + θP_{≥N}f)P_{≥N}f]∥_{L^{q}(R^{d})} ≲ ∥F′(P_{<N}f + θP_{≥N}f)P_{≥N}f∥_{L^{q}(R^{d})}

≲ ∥|P_{<N}f + θP_{≥N}f|^{p−1}|P_{≥N}f|∥_{L^{q}(R^{d})}

≲ ∥P_{<N}f + θP_{≥N}f∥_{L^{t}(R^{d})}^{p−1}∥P_{≥N}f∥_{L^{r}(R^{d})}

≲ ∥f∥_{L^{t}(R^{d})}^{p−1} ∑_{M≥N} ∥P_{M}f∥_{L^{r}(R^{d})}

≲ ∥f∥_{W^{s,r}(R^{d})}^{p−1}N^{−s}∥f∥_{W^{s,r}(R^{d})}

as desired.

Now we turn to (8). One could basically repeat the above analysis, but let us use a trick to deduce (8) from (a vector-valued version of) (7). We first use the fundamental theorem of calculus to write

F (f) − F (g) = ∫_{0}^{1} \frac{d}{dθ} F ((1 − θ)f + θg) dθ = ∫_{0}^{1} F′((1 − θ)f + θg)(f − g) dθ

and so by Minkowski it suffices to show that

∥P_{N}(F′((1−θ)f+θg)(f−g))∥_{W^{s,q}(R^{d})} ≲ (∥f∥_{W^{s,r}(R^{d})}+∥g∥_{W^{s,r}(R^{d})})^{p−1}∥f−g∥_{W^{s,r}(R^{d})}.

If we write f_{1} := (1 − θ)f + θg and f_{2} := f − g, it suffices to show that

∥P_{N}(F′(f_{1})f_{2})∥_{W^{s,q}(R^{d})} ≲ ∥f_{1}∥_{W^{s,r}(R^{d})}^{p−1}∥f_{2}∥_{W^{s,r}(R^{d})}.

We will now forget where f_{1}, f_{2} came from, and prove this estimate for all f_{1}, f_{2}. This allows us to perform the following normalisations. Firstly, as this estimate is homogeneous in f_{2} we may normalise ∥f_{2}∥_{W^{s,r}(R^{d})} = 1. Secondly, by replacing f_{1} with f_{1}/λ and F (x) with λ^{p}F (λx) (noting that this does not affect the hypotheses on F) we may also normalise ∥f_{1}∥_{W^{s,r}(R^{d})} = 1. Our goal is now to show that

∥P_{N}(F′(f_{1})f_{2})∥_{W^{s,q}(R^{d})} ≲ 1

when ∥f_{1}∥_{W^{s,r}(R^{d})} = ∥f_{2}∥_{W^{s,r}(R^{d})} = 1. Now for the key trick: we concatenate the two functions f_{1}, f_{2} into a single (vector-valued) function f_{12} := (f_{1}, f_{2}), and view the expression F′(f_{1})f_{2} as a function G(f_{12}) of the concatenated function, thus G(x, y) := F′(x)y. Observe that if F obeys the C^{2} bounds of the problem, then G obeys the C^{1} bounds. Thus we see now that (8) follows from the (vector valued) version of (7).

Now let us briefly sketch how one would prove the full Leibnitz rule. At this point we need to tap some more nontrivial results from harmonic analysis. We first need the Hardy-Littlewood maximal function

Mf(x) := sup_{r>0} \frac{1}{|B(x, r)|} ∫_{B(x,r)} |f(y)| dy

where B(x, r) is the ball of radius r centred at x. A standard fact from harmonic analysis is the Hardy-Littlewood maximal inequality

∥Mf∥_{L^{q}(R^{d})} ≲_{q} ∥f∥_{L^{q}(R^{d})}

for all 1 < q ≤ ∞. Less well known is the vector-valued generalisation of this, the Fefferman-Stein maximal inequality

∥(∑_{n} |Mf_{n}|^{r})^{1/r}∥_{L^{q}(R^{d})} ≲_{q,r} ∥(∑_{n} |f_{n}|^{r})^{1/r}∥_{L^{q}(R^{d})}

for any sequence of functions f_{n} and any 1 < q, r ≤ ∞. See e.g. Stein’s “Harmonic analysis”. The significance of the maximal function is that it controls all Littlewood-Paley projections pointwise:

|P_{N}f(x)|, |P_{<N}f(x)|, |P_{≥N}f(x)| ≲_{d} Mf(x).

For similar reasons it handles derivatives without difficulty, e.g.

|⟨∇⟩^{s}P_{N}f(x)| ≲_{d,s} ⟨N⟩^{s}MP_{N}f(x); |P_{N}f(x)| ≲_{d} N^{−1}M|∇P_{N}f(x)|.

We leave these as exercises.

The point is that we need to repeat the above analysis but at a pointwise level rather that a norm level, so that we can take square functions properly. Let us just prove the bounds for F (f), as the bounds for F (f) − F (g) follow from the trick outlined above. We split

∥F (f)∥_{W^{s,q}(R^{d})} ≤ ∥P_{<1}F (f)∥_{W^{s,q}(R^{d})} + ∥P_{≥1}F (f)∥_{W^{s,q}(R^{d})}.

The first part can be estimated as before. For the second part, we use the Littlewood-Paley inequality and reduce to showing that

∥( ∑_{N≥1} |P_{N}⟨∇⟩^{s}F (f)|^{2})^{1/2}∥_{L^{q}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}^{p}.

Now we argue as before and reduce to showing that

∥( ∑_{N≥1} |P_{N}⟨∇⟩^{s}F (P_{<N}f)|^{2})^{1/2}∥_{L^{q}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}^{p}.

and

∥( ∑_{N≥1} |P_{N}⟨∇⟩^{s}[F′(P_{<N}f + θP_{≥N}f)P_{≥N}f|]|^{2})^{1/2}∥_{L^{q}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}^{p}.

Let’s look at the first inequality. We have the pointwise bound

|P_{N}⟨∇⟩^{s}F (P_{<N}f)| ≲ N^{s−1}M|∇F (P_{<N}f)|
≲ N^{s−1}M[|F′(P_{<N}f)||∇P_{<N}f|]
≲ N^{s−1}M[|P_{<N}f|^{p−1}|∇P_{<N}f|].

We insert this bound and apply Fefferman-Stein to eliminate M, and reduce to computing

∥( ∑_{N≥1}[N^{s−1}|P_{<N}f|^{p−1}|∇P_{<N}f|]^{2})^{1/2}∥_{L^{q}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}^{p}.

We estimate P_{<N}f by Mf so that we can pull it out of the square function, reducing to

∥|Mf|^{p−1}( ∑_{N≥1}[N^{s−1}|∇P_{<N}f|]^{2})^{1/2}∥_{L^{q}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}^{p}.

We use Hölder (placing Mf in L^{t} and the square function in L^{r}). The Mf term can be handled by Hardy-Littlewood followed by Sobolev, and we reduce to

∥( ∑_{N≥1}[N^{s−1}|∇P_{<N}f|]^{2})^{1/2}∥_{L^{r}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}.

If f was band-limited to the ball {|ξ| ≤ 100} (say) then this is very easy, so by decomposition we can assume that f is instead band-limited to the exterior region {|ξ| ≥ 10} (say). Now the intuition is that the dominant portion of ∇P_{<N}f occurs at frequencies ∼ N (we already saw this in the preceding computations); to exploit this we can write P_{<N}f = ∑_{n≥0} P_{2^{−n}N}f and use the triangle inequality and reduce to

∑_{n≥0} ∥( ∑_{N≥2^{n}}[N^{s−1}|∇P_{2^{−n}N}f|]^{2})^{1/2}∥_{L^{r}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})};

here we use our previous reduction that the frequencies of f are ≥ 10. We estimate |∇P_{2^{−n}N}f| ≲ (2^{−n}N)^{1−s}M|P_{2^{−n}N}⟨∇⟩^{s}f| and apply Fefferman-Stein to reduce to showing

∑_{n≥0} 2^{−(1−s)n}∥( ∑_{N≥2^{n}}[|P_{2^{−n}N}⟨∇⟩^{s}f|]^{2})^{1/2}∥_{L^{r}(R^{d})} ≲ ∥f∥_{W^{s,r}(R^{d})}.

Applying the Littlewood-Paley inequality we estimate the left-hand side as

∑_{n≥0} 2^{−(1−s)n}∥f∥_{W^{s,r}(R^{d})}

and the claim follows.

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

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