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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!

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  • $\begingroup$ does this answer help at all? Or is there something you would still like clarified? $\endgroup$
    – Chill2Macht
    Apr 2, 2016 at 23:03

1 Answer 1

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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.

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    $\begingroup$ Thanks! But still it is not clear to me how the solution goes. Suppose you want to show that $\sum a_k\cos kx$ is convergent. Is your approach based on assuming that $\sum a_k$ is divergent? Also, I don't see why from the divergence of $\sum a_k$ you can conclude that $\sum a_k$ has unbounded partial sums. Are you assuming that $a_k\ge 0$ for all $k$? $\endgroup$
    – Simon
    Apr 3, 2016 at 3:07
  • $\begingroup$ if it diverges, that means by the definition of divergence that it goes to either plus or minus infinity. hence the sums are unbounded. it is not possible to diverge to a finite, hence bounded, value $\endgroup$
    – Chill2Macht
    Apr 3, 2016 at 3:51
  • $\begingroup$ en.wikipedia.org/wiki/Divergent_series; en.wikipedia.org/wiki/Lebesgue_integration $\endgroup$
    – Chill2Macht
    Apr 3, 2016 at 3:52
  • $\begingroup$ what we are trying to prove is: IF Fourier series converges, THEN both $\Sigma a_k$ and $\Sigma b_k$ converge. I showed that the contrapositive (which is logically equivalent) of this is true (IF one of $\Sigma a_k$ or $\Sigma b_k$ does not converge, ie diverges, ie has unbounded partial sums, then the Fourier series does NOT converge). $\endgroup$
    – Chill2Macht
    Apr 3, 2016 at 3:55
  • $\begingroup$ I'm still not convinced. Let $a_n=(-1)^n$. The $\sum a_n$ is divergent , but the partial sums are bounded. I hope it clarifies my objection to your solution. Also note that we want to show that $\sum a_n\cos nx$ is convergent, but it seems that you are trying to prove the convergence of $\sum a_n$. $\endgroup$
    – Simon
    Apr 3, 2016 at 3:56

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