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?

  • $\begingroup$ By converge do you mean "converge for all $t$"? Or is it just one $t?$ $\endgroup$
    – zhw.
    Mar 14, 2017 at 19:09
  • $\begingroup$ @zhw. For an arbitrary real $t$. In the end what I would like is the equivalence of $(1)$ and $(2)$ in the sense that for all real $t$, the series in $(1)$ converges if and only if the series in $(2)$ both converge and, in that case, the values are equal. $\endgroup$ Mar 14, 2017 at 19:12

3 Answers 3


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.

  1. 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, $f_{e}=\frac{f(x)+f(-x)}{2},\ f_{o}=\frac{f(x)-f(-x)}{2}$.

  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.

  3. 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.


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 $\sum_{k=1}^{\infty} [a_k \cos(kt) + b_k \sin(kt)]$.


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