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