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My question is in regards of Stein's proof that Hilbert transform is of weak $(1,1)$ property, on page 30 of the textbook I mentioned in my title.

On page 32 he writes that because $|\nabla K| \leq B |x|^{-n-1}$ where $B>0$ is some constant and $K$ is our kernel, that: $$|K(x-y)-K(x-y^j)| \leq B \frac{diameter(Q_j)}{|x-\bar{y}^j|^{n+1}}$$ where $y^j$ is the centre of the cube $Q_j$, and $\bar{y}^j$ is a variable point on the straight line segment connecting $y^j$ with $y\in Q_j$.

I am not sure I understand the last ineqaulity, I mean I know that $\Delta K = \nabla K \cdot \Delta \bar{x}$, where $\Delta\bar{x}$ is some change vector.

I am not sure I understand the term $|x-\bar{y}^j|$, obviously it comes from the upper bound for $\nabla K$, not sure how exactly.

Thanks in advance.

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1 Answer 1

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By the mean value theorem $$ K(x-y)-K(x-y^j)=\nabla K(z)\cdot(y-y^j), $$ where $z$ is a point between $x-y$ and $x-y^j$, that is, $z=x-\bar y^j$ where $\bar y^j$ is a point (depending on $x$) in the line segment from $y^j$ to $y$ ; the point may be different for different values of $x$. Then, since $y,y^j\in Q_j$ and $|\nabla K(x)|\le B\,|x|^{-n-1}$: $$ |K(x-y)-K(x-y^j)|\le|\nabla K(x-\bar y^j)|\,|y-y^j|\le\frac{B}{|x-\bar y^j|^{n+1}}\operatorname{diam}(Q_j). $$

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Thanks, that clears it for me. –  MathematicalPhysicist Apr 20 '12 at 6:42

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