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!

niceoperation: 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 9:44easyto 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