Estimate of a convolution from a paper by Michael Christ I don't understand Lemma 2 of the paper Hilbert transforms along curves, II: A flat case by Michael Christ. The situation is as follows. I slightly simplified it from the exact context in the source.
Let $\zeta\in C_0^\infty([-1,1])$ and $\zeta\equiv 1$ on $[-1/2,1/2]$. Let $$a_k(\xi,\eta)=\zeta(2^{-k}\xi)\zeta(2^{-2k}\eta),$$
$$b_k=a_k-a_{k-1}.$$

Lemma. Let $f$ be supported on a rectangle $R$ of dimensions $(2^{-k},2^{-i})$  where $k\ge 0$ and $i\in \{2k,2k-1\}$ and have mean value zero. Then we have for all $0\le j\le k$,
  $$\|f*\check{b}_j\|_1\le C 2^{j-k} \|f\|_1.$$

Here $\check{f}$ denotes the inverse Fourier transform of $f$.
Note that $i,j,k$ always denote integers here.
Christ says this Lemma follows in a "routine fashion" from Taylor's theorem and upper bounds for the derivatives of $b_j$. He doesn't give any more details.
But I haven't been able to do it. I tried sneaking in a $\check{b}_j$ using mean value property of $f$ and then applying Taylor's formula, but it doesn't seem to work out right. Also I'm wondering if it matters that there is a $b_j$ in the Lemma or if the $a_j$ would actually work just as well.
Can you please help me?
 A: As daw already indicated
$$ f* \check{b}_j(x) = \int f(y) \check{b}_j(x -y) \mathrm{d}y  =\int f(y) [\check{b}_j(x-y) - \check{b}_j(x - y_0)] \mathrm{d}y $$
by the constraint on the mean value; here $y_0$ is some fixed point in the support of $f$. Observe further that from the fundamental theorem of calculus, let $\gamma$ be a piecewise $C^1$ path with endpoints on $-y_0$ and $-y$, we have
$$ \check{b}_j(x-y) - \check{b}_j(x - y_0) = \int_0^1 \dot{\gamma}(s) \cdot (\partial\check{b}_j)[x + \gamma(s)] ~ \mathrm{d}s $$
So expanding and swapping order of integration by Fubini we have
$$ \| f * \check{b}_j \|_1 \leq \int |f(y)| \int_0^1 \int |\dot{\gamma}(s)\cdot\partial \check{b}_j(x +\gamma(s))| \mathrm{d}x ~\mathrm{d}s ~ \mathrm{d}y $$
Now for each fixed $y$, we pick a path $\gamma$ that goes first vertically (which we call $\gamma_2$) and than horizontally (which we call $\gamma_1$). By assumption on the support of $f$ the vertical length of $\gamma$ is at most $2 * 2^{-2k}$. Over that portion $\dot{\gamma}\cdot \partial \check{b}_j$ is taking the partial derivative relative to the second component. And on the horizontal portion we know that by assumption the length is at most $2^{-k}$ and the derivative is the partial derivative relative to the first component. 
So we have that 
$$ \| f * \check{b}_j \|_1 \leq \int |f(y)| \left[ 2\cdot 2^{-2k} \int |\partial_2\hat{b}_j(x) |~\mathrm{d}x + 2^{-k} \int| \partial_1 \hat{b}_j(x)| ~\mathrm{d}x \right] \mathrm{d}y $$
Since $b$ is $C^\infty_0$ we have that its Fourier transform is $\mathcal{S}$, so that $\| \partial \check{b}_j \|_1$ is well defined; in particular we have that $\| \partial\check{b}_1\|_1$ is just some constant. Now, we have that using the scaling law for the Fourier transform that
$$ \partial_1 \hat{b}_j(u,v) = 2^j 2^j 2^{2j} \partial_1\hat{b}_1(2^{j}u, 2^{2j}v) $$
and
$$ \partial_2 \hat{b}_j(u,v) = 2^{2j} 2^j 2^{2j} \partial_2 \hat{b}_1(2^{j}u, 2^{2j}v).$$
So we have that, using the Jacobian change of variables formula,
$$ \| \partial_1 \check{b}_j \|_1 = 2^{j} \|\partial_1 \check{b}_1 \|_1 $$
and 
$$ \|\partial_2 \check{b}_j \|_1 = 2^{2j} \|\partial_2 \check{b}_1\|_1 $$
This implies
$$ \|f * \check{b}_j \|_1 \leq \|f\|_1 \left[ 2 \cdot 2^{2j - 2k} \|\partial_2 \check{b}_1\|_1 + 2^{j-k} \|\partial_1 \check{b}_1\|_1 \right] $$
So the desired inequality holds with 
$$ C = 2 \| \partial_2 \check{b}_1 \|_1 + \|\partial_1 \check{b}_1\|_1 \geq 2^{j-k+1} \|\partial_2 \check{b}_1 \|_1 + \|\partial_1 \check{b}_1\|_1 $$
using that $j \leq k$ by assumption. 
A: Here is a hint how to deal with the convolution. I have no idea how to put in the inverse Fourier transform of the $b_j$.
All integrals are over $\mathbb R^2$. Then we have
$$
\int f\ast b\ dx= \int\int f(x-y)b(y)\ dy\ dx = \int\int ( f(x-y)-f(x))b(y)\ dy\ dx\\
=  \int\int f(x) (b(y-x)-b(y)).
$$
The third integral is derived using that mean value of $f$ is zero.
To obtain the fourth integral I used a coordinate transform to get from $f(x-y)b(y)$ to $f(z)b(x-z)$, then replacing the letters $(z,x)\to (x,y)$.
Now you use the fact that the support of $f$ is bounded, then the integrand is not zero only for $|x_1|\le 2^{-k}$ and $|x_2|\le 2^{-i}$. 
The difference $b(y-x)-b(y)$ can then be estimated using Taylor theorem.
Hope this helps.
