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I am struggling with the following.

Let $f$ be a $C^1$ function with period $2\pi$ and Fourier coefficients $c_n$. Prove that

  1. if $f$ is in $C^k$, then $|c_nn^k|\leq M_k$, where $M_k$ is a constant independent of $n$.
  2. if $\sum_{n=-\infty}^{\infty}|c_nn^k|$ is convergent, then $f$ is in $C^k$.

For the first question, I am not sure if I can just say that $c_n(u^{(k)})=i^kn^kc_n(u)$.

For the second part, I have no idea.

Thanks for the help.

Eman Yalpsid
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Mr. Tea
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1 Answers1

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To set the notation, let us define the Fourier coefficients by the following integral \begin{align} c_n = \frac{1}{2\pi}\int^{\pi}_{-\pi} f(x)e^{-in x}\ dx \ \ \ \forall \ n \in \mathbb{Z}. \end{align} Since $f$ is $2\pi$-periodic, then we have that \begin{align} c_n =&\ \frac{-1}{2\pi in}\int^{\pi}_{-\pi} f(x)\frac{d}{dx}e^{-in x}\ dx = \frac{-1}{2\pi in}f(x)e^{-inx}\Big|^{\pi}_{-\pi} + \frac{1}{2\pi i n}\int^\pi_{-\pi}f'(x)e^{-inx}\ dx\\ =&\ \frac{1}{2\pi in} \int^{\pi}_{-\pi}f'(x) e^{-inx}\ dx. \end{align} Repeat the process $k$-times, we get \begin{align} c_n = \frac{1}{2\pi (in)^k} \int^\pi_{-\pi} f^{(k)}(x)e^{-inx}\ dx. \end{align} Hence it following \begin{align} |c_n| \leq \frac{1}{2\pi n^k} \int^{\pi}_{-\pi}|f^{(k)}(x)|\ dx = \frac{M_k}{n^k} \ \ \Rightarrow \ \ |c_n|n^k \leq M_k. \end{align}

For the second question, let us begin by assuming \begin{align} \sum^\infty_{n=-\infty} |c_n|n^k<\infty. \end{align} By the hypothesis, we know that $f \in C^1$ and $2\pi$-periodic. Using the classical result of uniform convergence of Fourier series, then we know that \begin{align} f(x) = \sum^\infty_{n=-\infty}c_ne^{inx} \end{align} for all $x$. Lastly, observe \begin{align} f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0}\sum^\infty_{n=-\infty} c_n\frac{e^{-in(x+h)}-e^{-inx}}{h} =\sum^\infty_{n=-\infty} (-inc_n) e^{-inx} \ \ (\ast) \end{align} where the validity of interchanging limits in the last inequality comes from the uniform converges of the Fourier series. One should note the right-hand side series of $(\ast)$ converges uniformly since \begin{align} \sum^\infty_{n=-\infty}|c_n|n^k<\infty. \end{align}

Repeat the process $k$ times will show that $f \in C^k$.

Jacky Chong
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