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Let $X$ be a compact topological space, and $f: S^1 \times X \rightarrow \mathbb{C}$ be a continuous function. Then for every point $x \in X$ we can compute the Fourier coefficients $$c_k(x) = \int_{S^1} f(z,x) z^k \text{d}z,$$ and construct the Fourier series as the limit of the partial sums $$s_n(z, x) = \sum_{k=-n}^n c_k z^k.$$ This must not converge point-wise, but taking Cesàro means $$ \sigma_l(z,x) = \frac 1 l\sum_{n = 0}^{l-1} s_n(z,x) = \frac 1 l \sum_{n=0}^{l-1}\sum_{k=-n}^{n}c_kz^k = \sum_{k=-l}^l\frac{l - |k|}{l}c_kz^k,$$ we get uniform convergence in $z$ by Fejér's theorem.

Note that $\sigma_l(z,x)$ is still continuous in $x$.

In his book K-Theory, Atiyah claims that the $\sigma_l$ converge uniformly in $x$, that is given $\epsilon > 0$ there exists $l_0 \in \mathbb{N}$, such that for all $(z,x) \in S^1 \times X$ and $l \geq l_0$ we have $$\|f(z,x) - \sigma_l(z,x)\| < \epsilon .$$

I don't have any idea why this should be true, but my knowledge about Fourier series and analysis is pretty limited. Any help would be appreciated.

Here are some thoughts I had:

  • One obstacle is that the $\sigma_l$ do not form a true series, because of the factor $\frac 1 l$. So the classical Weierstrass test does not even apply here. Is there another kind of Weierstrass test that one could apply to this situation?

  • Because $X$ is compact, it would suffice to show uniform convergence locally.

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