User matthew - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-19T05:50:58Zhttp://mathoverflow.net/feeds/user/12885http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/55046/differentiating-spherical-harmonic-expansions-w-r-t-an-auxiliary-variableDifferentiating spherical harmonic expansions w.r.t. an auxiliary variableMatthew2011-02-10T15:54:06Z2011-02-10T15:54:06Z
<p>Consider a function <code>$f(x,\theta,\phi) \in C^{\infty}_0(\Re \times S^2)$</code>. For fixed $x\in \Re$, it's well known that the spherical harmonic decomposition <code>$f(x,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{|m|\leq l} f^{lm}(x)Y_{lm}(\theta,\phi)$</code>
converges uniformly and absolutely with respect to $\theta,\phi$, where $f^{lm} = \langle f(x,\cdot,\cdot),Y_{lm}\rangle_{L^2(S^2)}$ and $Y_{lm}$ are the spherical harmonics. However, I'm wondering if I can compute $f_x$ via term-by-term differentiation. More specifically, can we say that
<code>$f_{x}(x,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{|m|\leq l} f^{lm}_x(x)Y_{lm}(\theta,\phi)$</code>. This seems fairly standard, but my normal tool (DCT) for doing this doesn't help here. Is there a reference that contains this or an easy way to prove it?</p>
<p>Thanks!</p>