Given $$\sum_{n=1}^\infty\frac{\cos nt}{n}$$is it a fourier series in

a. $L^2(\mathbb T)$?

b. $C(\mathbb{T})$?

Usually when we get a series we use Weierstrass M test in order to find the sum is uniformly convergent, hence the function to which the series converges is continuous. Here, this test won't work since $$\sum_{n=1}^\infty\bigg|\frac{\cos nt}{n}\bigg|\le \sum_n\frac 1 n=\infty$$ We notice that the fourier coefficients (if it's a fourier series) are defined by $c_k=\frac{1}{2k}$.

How can we prove that in this case the series converges to a function ? Suppose we did so and not by uniform convergence: how would we disprove the continuity of the function?



  1. $\mathcal F: L^2(\mathbb T) \to \ell^2(\mathbb Z)$ is an isometry if defined via the complex fourier transform. Can you see that the complex fourier coefficients are in $\ell^2(\mathbb Z)$?
  2. Evaluate the series at $t=0$. Can this be continuous?
  • $\begingroup$ 1. Yes. thanks. 2. It that enough that in one point the function diverges to prove the function is not continuous? $\endgroup$ – user65985 Feb 11 '15 at 9:45
  • $\begingroup$ @DanisFischer Yes, because continuous functions on compact sets are bounded. $\endgroup$ – AlexR Feb 11 '15 at 9:45
  • $\begingroup$ Makes sense. thanks again. Last question: what if I was given a space which is not compact? $\endgroup$ – user65985 Feb 11 '15 at 9:47
  • $\begingroup$ Picking a compact subset the same argument applies. This will unlikely occur to you in this context, though. $\endgroup$ – AlexR Feb 11 '15 at 9:55
  • $\begingroup$ But if we can't? (For example take $\mathbb{R}$ with $\tau_{disc}$, which doesn't contain any compact subspace)? $\endgroup$ – user65985 Feb 11 '15 at 10:20

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