Uniform convergence of a trigonometric series

Question

Let $L>0$ be a constant. With what coefficients $\alpha_k$ and $\beta_k$ does the trigonometric series $$ \alpha_0 +\sum_{k=1}^{\infty} \left[\alpha_k \cos\left( \frac{k\pi x}{L} \right) +\beta_k \sin\left(\frac{k\pi x}{L}\right)\right] $$ converge uniformly?

Using the triangle inequality we have

$$ \left|\alpha_k\cos\left(\frac{k\pi x}{L}\right) + \beta_k\sin\left(\frac{k\pi x}{L}\right)\right| \le \left|\alpha_k\cos\left(\frac{k\pi x}{L}\right)\right| + \left|\beta_k\sin\left(\frac{k\pi x}{L}\right)\right| \le |\alpha_k| + |\beta_k|. $$

Now by the Weierstrass criterion the trigonometric series converges uniformly if the series $$\sum_{k=1}^{\infty}\left(|\alpha_k|+|\beta_k|\right)$$ converges.

Is there anything we can say about the coefficients $\alpha_k$ and $\beta_k$ other than the fact that $|\alpha_k|+|\beta_k|$ must approach zero as $k$ approaches infinity?

Answer 1

The sequence has to be convergent in l2 (so called Parserval's theorem) and not much more can be said about it. In fact, the following theorem is true: given a sequence {a_k} in l2. There exists continuous function with Fourier coefficients {b_k} such that |b_k| >= |a_k|. This result is due to S. Kisliakov I think.

Answer 2

One more comment, if uniform convergence is replaced with point-wise convergence almost everywhere, then the condition on Fourier coefficients {a_k} in l2 in sufficient. I.e. this is the famous Carlson-Hunt theorem. Examples of continuous functions with Fourier series not convergent uniformly are plenty and well known. One way to derive this is from estimates on norm of projection from continuous functions to polynomials of degree <=n (~Log n ) or construct such function directly (see Zygmund's texbook for example).