The spherical harmonics are given by $$Y^m_l(\phi,\theta):=N^m_l e^{im\theta}P^m_l(\cos \phi) $$ where $P^m_l$ are the associated Legendre Polynomials and $N^m_l$ is the normalisation. From inspecting the first few it seems $$\left\Vert Y^m_l \right\Vert \leq C <\infty.$$ Is this true? If not is there an easy upper bounded of the form $$\left\Vert Y^m_l \right\Vert\leq C^m_l <\infty.$$
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Which normalization are you using? $\endgroup$– Robert IsraelCommented Jun 25, 2015 at 16:39
-
$\begingroup$ Related questions (without answering this one: mathoverflow.net/questions/118692/…, mathoverflow.net/questions/55235/… $\endgroup$– András BátkaiCommented Jun 25, 2015 at 16:55
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
As mentioned above, your question cannot be answered without knowledge of which norming sequence you are using. However, there are uniform bounds for the derivatives of the Legendre polynomials in the literature (see, e.g., pp. 523, 524 of the article reviewed in MR 0306897 by Guillemont-Tessier, which is available on line) and these immediately give estimates for the functions you are interested in.
$\begingroup$
$\endgroup$
There is the general fact that when a compact topological group acts (continuously) transitively on a topological space with an invariant probability measure, a stable finite-dimensional vector space of functions has a sharp estimate for the ratio of sup-norm to $L^2$ norm: square root of the dimension of the space of functions.
(It is my impression that examination of the argument for that standard result shows that a function invariant under the isometry group of a point hits the bound.)
answered Jul 1, 2015 at 11:25