For reference, I am reading the paper "Uniqueness of Finite Energy Solutions for Maxwell-Dirac and Maxwell-Klein-Gordon Equations" by Masmoudi and Nakanishi.

Let $A_0$ be a scalar function satisfying the Poisson equation

$$\Delta A_0 = -|u|^2$$

where $u$ is taken from the space $X^{1/2,b} = U(t)H^b_t H^{1/2}_x + U(-t)H^b_t H^{1/2}_x$ (formally) where $U(\pm t) = \exp(\pm i|\nabla|t)$ is the free wave propagator.

The authors claim the following estimate holds true $$ \|A_0\|_{L^{\infty}(\dot{B}^{3/2}_{2,1})} \lesssim \| |u|^2 \|_{L^{\infty}(\dot{B}^{-1/2}_{2,1})} \lesssim \| u\|^2_{L^{\infty}H^{1/2}} $$

Here, $\dot{B}$ denotes the usual homogeneous Besov space.

I have two questions:

  1. Why do we lose two derivatives in the first estimate. Shouldn't we gain two? Or am I just confused here.
  2. I completely fail to understand the second estimate. Is there an embedding $H^{1/2} \hookrightarrow \dot{B}^{-1/2}_{2,1}$? The only thing I know is that $H^m$ is isomorphic to $B^m_{22}$ for $m\in \mathbb{N}$ but here we have non-integer $s$.

EDIT: As described in the comments under this post, I changed the last term in the chain of inequalities.

  • $\begingroup$ Are you sure you copied the final inequality correctly? In particular, are you sure it is $|u|^2$ being measured in $L^\infty H^{1/2}$, and not $u$ (with the square going, perhaps, on the outside)? As stated the second inequality is clearly not true. $\endgroup$ Oct 9, 2019 at 15:47
  • $\begingroup$ @WillieWong Thank you for pointing out my mistake, the last term should of course be as you described. I will edit this. $\endgroup$ Oct 9, 2019 at 18:43

2 Answers 2


Here's to address the second inequality.

First, I am pretty certain you copied the inequality wrong, and it should read (dropping the $L^\infty$ factor in time)

$$ \| |u|^2 \|_{\dot{B}^{-1/2}_{2,1}} \lesssim \|u\|_{H^{1/2}}^2 $$

Second, I am a bit amused when I glanced at the original paper that I can't find anywhere obvious where they specified a crucial piece of information, namely that they are working on spatial dimension 3. (It is not in the abstract nor on the first two pages of the introduction, as far as I can see.)

So here goes the usual dichotomy argument. Write $u = \sum_{k} P_k u$ where $P_k$ is the Littlewood Paley projector to frequency $\approx 2^{k}$, here $k$ runs over $\mathbb{Z}$.

The Besov norm we can write as $$ \sum_{k} 2^{-k/2} \| P_k (\sum_j P_j u)(\sum_\ell P_\ell u) \|_{L^2} $$ The product properties of Littlewood Paley projectors (since Fourier transform moves products to convolutions), shows that

  1. When $j > \ell + 1$, then $P_j u P_\ell u$ has frequency support approximately at $2^{j}$
  2. When $j = \ell, \ell + 1, \ell - 1$, then $P_j u P_\ell u$ has frequency support over roughly the entire ball of radius $2^{j}$.

So we can expand (roughly speaking) the Littlewood Paley sums to read

$$ \lesssim \sum_k 2^{-k/2} \left( \| \sum_{j > k} P_k (P_j u)^2 \|_{L^2} + \| \sum_{j \leq k} (P_j u) (P_k u) \|_{L^2} \right) $$

First Term

We use the Sobolev embedding $\dot{W}^{3/2,1} \hookrightarrow L^2$ (in 3 dimensions) to get

$$ \|P_k (P_j u)^2 \|_{L^2} \lesssim 2^{3k/2} \| P_k (P_j u)^2 \|_{L^1} $$

Next we drop the projection and get

$$ \lesssim 2^{3k/2} \| P_j u\|_{L^2}^2 $$

So the first term is bounded by

$$ \text{First term} \lesssim \sum_{k} 2^{k} \sum_{j > k} \|P_j u\|_{L^2}^2 = \sum_j \|P_j u\|_{L^2}^2 \sum_{k < j} 2^{k} \leq 2 \sum_j 2^{j} \|P_j u\|_{L^2}^2 \lesssim \|u\|_{H^{1/2}}^2 $$

Second term

Start with

$$ \sum_{j \leq k} \|P_{j} u P_k u\|_{L^2} \lesssim \sum_{j \leq k} \|P_{j} u\|_{L^\infty} \|P_k u\|_{L^2} $$

Bernstein's inequality (or the frequency restricted Sobolev inequality, which holds also in the case of $L^\infty$) in 3 dimensions implies

$$ \lesssim \sum_{j \leq k} 2^{3j/2} \|P_j u\| \|P_k u\| $$

So we can write (symmetrizing in $j$ and $k$)

$$ \text{Second term} \lesssim \sum_{j} \sum_{k} 2^{- |j-k|} 2^{j/2} \| P_j u\|_{L^2} 2^{k/2} \|P_k u\|_{L^2} $$

So by Cauchy-Schwarz with weights, we have

$$ \lesssim (\sum_{j,k} 2^{-|j-k|} 2^{j} \|P_j u\|_{L^2}^2 )^{1/2} (\sum_{j,k} 2^{-|j-k|} 2^{k} \|P_k u\|_{L^2}^2 )^{1/2} $$

The two sums are identical after swapping $j$ and $k$, and summing first over $j$ in the second one you get

$$ \sum_{j,k} 2^{-|j-k|} 2^{k} \|P_k u\|_{L^2}^2 = 3 \sum_k 2^{k} \|P_k u\|_{L^2}^2 $$

and so

$$ \text{Second term} \lesssim \sum_{k} 2^{k} \|P_k u\|_{L^2}^2 \lesssim \|u\|_{H^{1/2}}^2. $$

  • $\begingroup$ I think the reason that it is tacitly assumed that $n=3$ is that the work is on the Maxwell-Dirac equation which physically only makes sense in space dimension three. But thank you very much again for pointing out the mistake and for this concise answer. $\endgroup$ Oct 9, 2019 at 19:00

In the first estimate you gain, not lose, two derivatives: if $|u|^2$ is only $\dot B^{-1/2}$ then $A_0=- \Delta^{-1}|u|^2$ is two orders more regular, so it belongs to $\dot B^{3/2}$.

EDITED: Concerning the second estimate, I am not convinced about it. Indeed, one has always $\dot B^s_{2,1}\hookrightarrow\dot B^s_{2,2}=\dot H^s$ by the simple fact that $\ell^1 \hookrightarrow\ell^2$. Thus an estimate $H^{1/2}\hookrightarrow \dot B^{-1/2}_{2,1}$ would imply $H^{1/2}\hookrightarrow \dot H^{-1/2}$. But this is not true: take $v$ a test function which is equal to 1 in a nbd of 0 and define $u$ as the Fourier transform of $|\xi|^{(1-n)/2}v(\xi)$, $n$ being the space dimension; then $u\in H^{1/2}$ but $u\not\in \dot H^{-1/2}$. Thus it is not true that $H^{1/2}\hookrightarrow \dot B^{-1/2}_{2,1}$. Maybe other pieces of information about $|u|^2$ are used?

  • $\begingroup$ Thank you. Could you maybe elaborate on $B^{1/2}_{2,2} \hookrightarrow B^{1/2-\epsilon}_{2,1}$? Is it true that $B^{s}_{pq_1} \hookrightarrow B^{s-\epsilon}_{pq_2}$ for any two $1\leq q_1,q_2 \leq \infty$? $\endgroup$ Oct 9, 2019 at 11:18
  • $\begingroup$ Sure. My answer is not complete so let me edit it. I did not notice the dots! Feel free to de-approve my answer $\endgroup$ Oct 9, 2019 at 14:16
  • 2
    $\begingroup$ @PieroD'Ancona: I am pretty sure the OP copied the inequality wrong, and the second inequality should be a product estimate. (See my answer.) $\endgroup$ Oct 9, 2019 at 17:17
  • 1
    $\begingroup$ I have corrected the mistake that WillieWong was pointing out. $\endgroup$ Oct 9, 2019 at 18:46

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