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I don't want to write precisely the formulation of the Calderon-Zygmund theorem for singular integrals. The details are not so important here.

So I consider the operator $T$ given by the following formula: \begin{equation} Tf(x) = \int\limits_{R^n}\dfrac{Y_{k}^m(\frac{x-y}{|x-y|})}{|x-y|^n}f(y)\, dy, \, f\in L^2(R^n), \end{equation} where $Y_k^m$ is a spherical harmonic of degree $k$ ($m$ denotes the index in the basis of the related subspace of harmonics on $S^{n-1}$).

It is known that operator $T$ is well-defined in $L^2(R^n)$ (and some other spaces as well). In particular, the following estimate is usually given in the proofs of the Calderon-Zygmund theorem: \begin{equation} \|Tf\|_{L^2} \leq C(n)\|\mathcal{F}[\frac{Y_{k}^m}{r^n}]\|_{L^{\infty}} \|f\|_{L^2}, \end{equation} where $\mathcal{F}$ denotes the Fourier transform (maybe in the mean-value sense), $C(n)$ is just some constant depending only on dimension of space.

1) Is the latter estimate is precise in the sense that there is only $L^{\infty}$-norm of the Fourier transform of the kernel (of this particular kernel with spherical harmonics)? Can this norm, for example, be improved somehow (any $L^p$ norm)?

2) What if, for example, $f$ belongs to a better space, i.e., $f\in L^{\infty}(R^n)$, $\mathrm{supp}\, f\subset D$, where $D$ is some compact?

P.S. The point is that, for example, in $n=3$ (and in any dimension as well) the Fourier transform of such kernel is the spherical harmonic $Y_k^m$ (it is known fact), but the maximum of a spherical harmonic on a sphere grows as $\sim \sqrt{2k+1}$, which is ofcourse more than, for example, as an $L^2$-norm: $\|Y_{k}^m\|_{L^2(S^2)} =1$.

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  • $\begingroup$ If someone can present me a function $f$ that the given estimate precise I will be very happy also! $\endgroup$
    – Fedor Goncharov
    Commented Sep 10, 2017 at 16:50
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    $\begingroup$ This is perhaps a stupid question, but what is $r$ in the estimate for $\|Tf\|_{L^{2}}$? $\endgroup$
    – Matt Rosenzweig
    Commented Sep 10, 2017 at 19:25
  • $\begingroup$ Why not use the Young inequality for convolutions in the real space? $\endgroup$
    – timur
    Commented Sep 10, 2017 at 21:38

1 Answer 1

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Your operator $T$ is a Fourier multiplier with symbol $m = \mathcal{F}[Y^m_k/r^n]$; that is, $\mathcal{F}[Tu]=m \mathcal{F}[u]$.

It is a relatively simple exercise to show that the norm of $T$ on $L^2$ is equal to $\|m\|_\infty$, the essential supremum of $|m|$. In other words, $C(n)=1$. Therefore, the supremum norm of $m$ cannot be change to any other norm.

Calderón–Zygmund theory provides estimates of the norm of $T$ on other $L^p$ spaces (and much more).

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  • $\begingroup$ On L^2 -- yes, I thought of some function, whose Fourier transform is concentrated in directions, where the harmonic Y_{k}^m reaches it's maximum and go for the limit, compressing the cone of these directions. However, for the bounded functions it doesn't work, because the Fourier transform is analytic and by the Theorem of Uniqueness cannot be supported only in some small cone. But your answer is very helpful for me already! $\endgroup$
    – Fedor Goncharov
    Commented Sep 11, 2017 at 5:21
  • $\begingroup$ In the previous comment I've made a mistake -- not only bounded but also compactly supported. $\endgroup$
    – Fedor Goncharov
    Commented Sep 11, 2017 at 6:05
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    $\begingroup$ As long as you keep the $L^2$ norm of $f$ and $T f$, the estimate cannot be improved: even $C_c^\infty$ is dense in $L^2$, so you can approximate any $L^2$ function (in the $L^2$ sense) by smooth, compactly supported functions. So pick $g \in L^2$ so that $\|g\|_2=1$ and $\|Tg\|_2\geqslant\|m\|_\infty-\varepsilon$, then pick $f\in C_c^\infty$ so that $\|f\|_2=1$ and $\|f-g\|_2 \leqslant \varepsilon$, and observe that $\|T f\|_2 \geqslant \|Tg\|_2-\|T(f-g)\|_2 \geqslant \|m\|_\infty - \varepsilon - \|m\|_\infty \|f-g\|$ is close to $\|m\|_\infty$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 11, 2017 at 7:58
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    $\begingroup$ And if you like to keep the $L^2$ norm of $Tf$ and consider a different norm of $f$, then there is little hope that you can get any reasonable answer. At least for the $L^p$ norm $\|f\|_p$ there is no such bound: simply consider a rescaled version of $f$, $g(x) = f(k x)$, and note that both sides of the inequality $\|Tg\|_2 \leqslant C \|g\|_p$ are homogeneous in $k$ with a different exponent. Maybe you could get some improvement if you considered expressions of the form $\|f\|_1+\|f\|_\infty$ or $(\|f\|_1\|f\|_\infty)^{1/2}$ in the right-hand side? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 11, 2017 at 8:03
  • $\begingroup$ Yes, you are obviously right about the compact support, I missed that point. And the idea of using another norm is also good. I've just found that having the $L^p$-norm of the Fourier transform I still can do estimates of the function $f$ in $L^q$-norm (q- is the dual conjugate). en.wikipedia.org/wiki/Riesz%E2%80%93Thorin_theorem $\endgroup$
    – Fedor Goncharov
    Commented Sep 11, 2017 at 11:10

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