Spherical Harmonics expansion for $f\in L^{2}(S^{n-1})$. Let $f\in L^{2}(S^{n-1})$ where $S^{n-1}$ is the unit sphere of $\Bbb{R}^n$.
As it is known $f$ has the  following spherical Harmonics expansion (the convergence is $L^{2}(S^{n-1})$ in :
$f(w)=\sum_{j} Y_j(w)$ where $Y_j\in H_j$ with $H_j$ is the space of spherical harmonics of degree $j$. 
Now let $f\in C(S^{n-1})$ (space of continuous functions on S^{n-1} ) then  $f\in  L^{2}(S^{n-1})$.
Hence $f(w)=\sum_{j} Y_j(w)$.
My question can we say that the series in question converge uniformly and absolutely on $S^{n-1}$.
 A: No, such convergence fails already on $S^1$, in the theory of Fourier series, for example. And, already in the ordinary theory of Fourier series, there are at least two nice assumptions to achieve uniform pointwise convergence: absolute summability of the Fourier coefficients (e.g., from assuming $f$ is $C^1$), or, a Sobolev-type inequality, such as knowing $f$ and its distributional derivative $f'$ are both in $L^2$. But either sort of condition increases in severity as dimension $n$ of $(S^1)^n$ goes up: for the $L^2$ Sobolev inequality, we need more than $n/2$ distributional derivatives to be in $L^2$.
On higher-dimensional spheres, there is likewise a range of possible assumptions to guarantee uniform pointwise convergence, but $L^2$ is far from enough, and all the worse as dimension goes up. As on a product of circles, sufficient smoothness suffices, and/or $L^2$-ness of sufficiently many derivatives $\Delta^k f$ where (e.g.) $\Delta$ is the rotation-invariant Laplace-Beltrami operator on the sphere. The Sobolev index shift is again dimension/2.
