# Differentiating spherical harmonic expansions w.r.t. an auxiliary variable

Consider a function $f(x,\theta,\phi) \in C^{\infty}_0(\Re \times S^2)$. For fixed $x\in \Re$, it's well known that the spherical harmonic decomposition $f(x,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{|m|\leq l} f^{lm}(x)Y_{lm}(\theta,\phi)$ 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 $f_{x}(x,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{|m|\leq l} f^{lm}_x(x)Y_{lm}(\theta,\phi)$. 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?

Thanks!

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This is mostly guesswork since I don't know the detailed formulae for spherical harmonics. Am I right in guessing that this is a kind of orthogonal expansion, similar to Fourier series but with a different basis? As a first step, examine the differentiability properties of $f^{lm}$. Presumably the explicit formula, an integral over $S^2$ against $Y_{lm}$, can be differentiated directly? Hopefully, this should tell you something about the decay properties of the coefficients. If you can show that they decay fairly quickly as $l,m \to \infty$, then you will get absolute convergence... –  Zen Harper Feb 15 '11 at 9:35
...(continued) hopefully with respect to some nice norm. Assuming that the formal series for $f_x$ really does converge nicely (which is the hard part), it shouldn't be too hard to show term-by-term differentiation, by a standard TRICK: consider $\int \ldots dx$ on the formal series for $f_x$, and INTEGRATE term-by-term instead of differentiating. This should work nicely because integration is a nice operation: usually it's easier to show something like $\int_{a}^{b} \sum_j \frac{\partial g_j}{\partial x} dx = \sum_j\int_{a}^{b} \frac{\partial g_j}{\partial x} dx$. –  Zen Harper Feb 15 '11 at 9:44
...(continued again; more waffle!!) As an analogy, consider Fourier series; if $f: R \to R$ is periodic with Fourier expansion $f(x) = \sum_n a_n e^{inx}$, and is known to have nice differentiability (say $C^3$), then $a_n = O(1/n^3)$ by integration by parts (directly from the formula), so that $g(x) = \sum_n in a_n e^{inx}$ converges absolutely and uniformly. Now it is easy to show that $\int_0^x g(t) dt = f(x)-f(0)$, so that we must have $f'(x) = g(x)$ just by the easy half of the Fundamental Theorem of Calculus. But if you try to estimate $[f(x+h)-f(x)]/h$ directly then it's very messy! –  Zen Harper Feb 15 '11 at 9:53
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