Discuss whether or not it is possible to have a Fourier series $$a_0+\sum_{k=1}^\infty[a_k\cos(kx)+b_k\sin(kx)]$$ converge for all $x$ without either $$a_0+\sum_{k=1}^\infty a_k\cos(kx) \text{ or } \sum_{k=1}^\infty b_k\sin(kx)$$ converging.
This is a problem in Bressoud's analysis book and my solution is as follows: "No, because if we let $f(x)=a_0+\sum_{k=1}^\infty[a_k\cos(kx)+b_k\sin(kx)]$, then the two other series are obtained by taking $\frac{f(x)\pm f(-x)}{2}$ and since $f(x)$ is convergent for all $x$ the sine and cosine series should also be convergent."
Here is the hint from the back of the book:
If the Fourier series converges at $x=0$, then $\sum_{k=1}^\infty a_k$ converges, and therefore the partial sums of $\sum_{k=1}^\infty a_k$ are bounded.
Although I think my solution is correct (please correct me if I'm wrong) I still would like to see other solutions and in particular understand the author's hint since I can't see how the boundedness of partial sums can help.
Thanks!
1 Proceeding by contraposition, if one of those two series $\Sigma a_i$,$\Sigma b_i$ doesn't converge (that they are absolutely convergent follows from the fact that the Fourier series converges everywhere) i.e. equivalently if one of $\Sigma a_i$,$\Sigma b_i$ have unbounded partial sums, then that implies that the partial sums of the Fourier series itself must also be unbounded (sum of limits is limit of sum) which contradicts the assumption that the Fourier series converges.
I.e. the boundedness of partial sums that you mentioned is a necessary condition for series convergence, hence if it fails for either $\Sigma a_i$,$\Sigma b_i$, then it fails for the Fourier series as a whole, and hence the Fourier series does not converge at all x.
(note that this follows because of absolute convergence of the series and the monotone/dominated convergence theorem)
2 Your solution is also correct. For example, if $\frac{f(x)-f(-x)}{2}$ is not convergent, then either f(x) or f(-x) is not convergent, which contradicts the assumption that the Fourier series converges everywhere.