# Calderon-Zygmund theorem for the kernel of spherical harmonics

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: $\|Y_{k}^m\|_{L^2(S^2)} =1$.

• If someone can present me a function $f$ that the given estimate precise I will be very happy also! – Fedor Goncharov Sep 10 '17 at 16:50
• This is perhaps a stupid question, but what is $r$ in the estimate for $\|Tf\|_{L^{2}}$? – Matt Rosenzweig Sep 10 '17 at 19:25
• Why not use the Young inequality for convolutions in the real space? – timur Sep 10 '17 at 21:38

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).
• 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 $\|m\|_\infty$. – Mateusz Kwaśnicki Sep 11 '17 at 7:58
• 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? – Mateusz Kwaśnicki Sep 11 '17 at 8:03
• 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 – Fedor Goncharov Sep 11 '17 at 11:10