On the equivalence of representations of Fourier series Let $f : \Bbb{R} \to \Bbb{C}$ be a $2\pi$-periodic function such that
$$
\int_0^{2\pi} |f(t)| \,dt < \infty
$$
Define
$$
\hat{f}(k) := \frac{1}{2\pi} \int_0^{2\pi} f(t) e^{-i k t} \,dt
$$
The Fourier series of $f$ is then
$$
\sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikt} \tag{1}
$$
If we define
$$
a_k := \frac{1}{\pi} \int_0^{2\pi} f(t) \cos(kt) \,dt \quad (k \geq 1) \\
b_k := \frac{1}{\pi} \int_0^{2\pi} f(t) \sin(kt) \,dt \quad (k \geq 1)
$$
then the Fourier series of $f$ takes the form
$$
\hat{f}(0) + \sum_{k=1}^{\infty} a_k \cos(kt) + \sum_{k=1}^{\infty} b_k \sin(kt) \tag{2}
$$
In passing from $(1)$ to $(2)$, I have the following question (I'm led to think the answer is yes but haven't succeeded in proving it):

If
  $$
\sum_{n=1}^{\infty}\left\{\hat{f}(n)[\cos(nt)+i\sin(nt)] + \hat{f}(-n)[\cos(nt)-i\sin(nt)]\right\} = \sum_{k=1}^{\infty} [a_k \cos(kt) + b_k \sin(kt)]
$$
  converges then do all four series
  $$
\sum_{n=1}^{\infty}\hat{f}(n)[\cos(nt)+i\sin(nt)] \\ \sum_{n=1}^{\infty}\hat{f}(-n)[\cos(nt)-i\sin(nt)] \\
\sum_{k=1}^{\infty} a_k \cos(kt) \\
\sum_{k=1}^{\infty} b_k \sin(kt)
$$
  converge?

 A: We define the Fourier series description as $\mathfrak{F}$, and can be written as follows:
$$
\begin{aligned}
\mathfrak{F}(\hat{f}(k))&=
\sum_{k=-\infty}^{\infty}
\hat{f}(k)e^{ikt}
\\&=
\sum_{k=-\infty}^{-1}
\hat{f}(k)e^{ikt} 
+\hat{f}(0)+
\sum_{k=1}^{\infty}
\hat{f}(k)e^{ikt} 
\\&=
\mathfrak{F}(\hat{f}(k))_{-}
+\hat{f}(0)+
\mathfrak{F}(\hat{f}(k))_{+}
\end{aligned}
$$
where $\mathfrak{F}(\hat{f}(k))_{+}$ is summation in positive side and $\mathfrak{F}(\hat{f}(k))_{-}$ is negative side. These summations have a following relation:
$$
\mathfrak{F}(\hat{f}(k))_{-}=\mathfrak{F}(\hat{f}(-k))_{+}
$$
Hence these four series can be derived as follows:
$$
\begin{aligned}
\sum_{k=1}^{\infty}\hat{f}(k)[\cos(kt)+i\sin(kt)]
=& \mathfrak{F}(\hat{f}(k))_{+} \\ 
\sum_{k=1}^{\infty}\hat{f}(-k)[\cos(kt)-i\sin(kt)]
=& \mathfrak{F}(\hat{f}(-k))_{+} \\
\sum_{k=1}^{\infty} a_k \cos(kt) 
=&
\cfrac{1}{2}\biggl[
\mathfrak{F}(\hat{f}(k)) + \mathfrak{F}(\hat{f}(-k))
\biggr]-\hat{f}(0)\\
\sum_{k=1}^{\infty} b_k \sin(kt)
=&
\cfrac{1}{2}\biggl[
\mathfrak{F}(\hat{f}(k)) - \mathfrak{F}(\hat{f}(-k))
\biggr]
\end{aligned}
$$
where example of derivation:
$$
\begin{aligned}
\sum_{k=1}^{\infty} a_k \cos(kt) =&
\sum_{k=1}^{\infty}
\biggl[
\cfrac{1}{\pi}\int_{0}^{2\pi}f(t)\cos(kt)dt
\biggr]
\cos(kt) \\=&
\sum_{k=1}^{\infty}
\biggl[
\cfrac{1}{2\pi}
\int_{0}^{2\pi}f(t)(e^{ikt}+e^{-ikt})dt
\biggr]
\frac{e^{ikt}+e^{-ikt}}{2}\\=&
\cfrac{1}{2}
\sum_{k=1}^{\infty}
\bigl(
\hat{f}(-k)+\hat{f}(k)
\bigr)
(e^{ikt}+e^{-ikt}) \\=&
\cfrac{1}{2}
\Biggl[
\sum_{k=1}^{\infty} \hat{f}(-k)e^{ikt} +
\sum_{k=1}^{\infty} \hat{f}(-k)e^{-ikt} +
\sum_{k=1}^{\infty} \hat{f}(k)e^{ikt} +
\sum_{k=1}^{\infty} \hat{f}(k)e^{-ikt}
\Biggr] \\=&
\cfrac{1}{2}\biggl[
\mathfrak{F}(\hat{f}(-k))_{+} +
\mathfrak{F}(\hat{f}(k))_{-} +
\mathfrak{F}(\hat{f}(k))_{+} +
\mathfrak{F}(\hat{f}(-k))_{-}
\biggr] \\=&
\cfrac{1}{2}\biggl[
\mathfrak{F}(\hat{f}(k)) + \mathfrak{F}(\hat{f}(-k))
\biggr]-\hat{f}(0).
\end{aligned}
$$
Owing to the problem's description, the function has finite norm, and Parseval's identity shows that the norm of Fourier series and function $f(t)$ are corresponded.
$$
\begin{aligned}
\|\mathfrak{F}(\hat{f}(k))\|_{2}^{2}=&
\|\mathfrak{F}(\hat{f}(-k))\|_{2}^{2}\\=&
\frac{1}{2\pi}\int_{0}^{2\pi}|f(t)|^2dt < \infty
\end{aligned}
$$
Therefore the Fourier series is convergent so that $\|\mathfrak{F}(\hat{f}(-k))_{+}\|_{2}^2$ and $\|\mathfrak{F}(\hat{f}(-k))_{-}\|_{2}^2$ are finite, and eventually above four series are also converged.
A: *

*If $f=f_{e}+f_{o}$, $f_{e}$ is even and $f_{o}$ is odd, then
$$ f(x)=f_{e}(x)+f_{o}(x)$$
$$ f(-x)=f_{e}(x)-f_{o}(x) $$
Then, $f_{e}=\frac{f(x)+f(-x)}{2},\ f_{o}=\frac{f(x)-f(-x)}{2}$.

*If $f\in L^{1}([0,2\pi])$ then $f_{o},\ f_{e}\in L^{1}([0,2\pi])$. So, we have the fourier expansions:
$$ f(x)=a_{0}+\sum_{k=1}^{\infty}(a_{k}\cos(kx)+b_{k}\sin(kt)) $$
$$ f_{o}(x)=\sum_{k=1}^{\infty}b_{k}'\sin(kt) $$
$$ f_{e}(x)=a'_{0}+\sum_{k=1}^{\infty}a'_{k}\cos(kx) $$
Since $f=f_{o}+f_{e}$ and by the uniqueness of the fourier expansion we see that $a_{k}=a_{k}',\ b_{k}=b_{k}'$. So, $\sum_{k=1}^{\infty}a_{k}\cos(kt)$ and $\sum_{k=1}^{\infty}b_{k}\sin(kt)$ converges.

*Put $S=\sum_{n=1}^{\infty}\hat{f}(n)[\cos(nt)+i\sin(nt)],\ T=\sum_{k=1}^{\infty}(-b_{k}\cos(kx)+a_{k}\sin(kt))$. Since for $k>0$ $\hat{f}(k)=a_{k}-ib_{k}$, we have 
$$ S=\sum_{k=1}^{\infty}(a_{k}\cos(kx)+b_{k}\sin(kt))+iT$$
If $S$ converges for every $f\in L^{1}([0,2\pi])$, $T$ also converges for such $f$, but this article says that there are functions for wich $T$ does not converges.
A: An attempt:
From choosing $t=0$  we get that $\sum_{k=1}^{\infty} a_k$ converges. Using the dirichlet criterium for convergence, we can see that $\sum_{k=1}^{\infty} a_k \cos(kt)$ converges. Now $\sum_{k=1}^{\infty} b_k \sin(kt)$ has to converge as well, otherwise we would get a contradiction with convergence of $\sum_{k=1}^{\infty} [a_k \cos(kt) + b_k \sin(kt)]$.
