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Problem: Let $ f:[0,2\pi] \to \mathbb{C} $ be continuously differentiable s.t. $ f(0) \neq f(2 \pi) $. Prove that $n \hat{f}(n)-\frac{i}{2 \pi}[f(2 \pi)-f(0)] \underset{n \rightarrow \pm \infty}{\longrightarrow} 0$

Attempt:
We'll integrate by parts ( allowed since $ f $ is continuously differentiable ) and get
$\widehat{f}(n)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(x) e^{-i n x} d x=\frac{1}{2 \pi}\left[f(x) \frac{e^{-i n x}}{-i n}\right]_{0}^{2 \pi}-\frac{1}{2 \pi} \int_{0}^{2 \pi} f^{\prime}(x) \frac{e^{-i n x}}{-i n} d x=$ $=\frac{1}{2 \pi} \cdot \frac{i \cdot[f(2 \pi)-f(0)]}{n}+\frac{1}{i n} \cdot \widehat{f}^{\prime}(n)$
Hence if we multiply by $ n $ and transfer sides in the equation we get
$n \hat{f}(n)-\frac{i}{2 \pi}[f(2 \pi)-f(0)] = \frac{1}{i}\cdot \widehat{f}^{\prime}(n)$
[ missing arguments to continue ]

Context + Notes:
The exercise above is from my TA's notes, under the subject of Riemann-Lebesgue lemma which is stated in my course as:

Lemma ( Riemann-Lebesgue Lemma ): Let $ f \in R(\mathbb{T}) $. Thus $ \lim_{n \to \pm \infty} | \hat{f}(n) | = 0$

  • $ R(\mathbb{T}) $ is the set of $ 2\pi $ periodic and Riemann Integrable functions from $ [0, 2\pi] $

I've learned that

If $ f \in C^1({\mathbb{T}}) $ ( meaning $f$ is continuously differentiable and $ f(0) = f( 2 \pi ) $ ). Thus $ \hat{f}'(n) = in \cdot \hat{f}(n) $

( But I can't use this in the exercise above since $ f(0) \neq f( 2 \pi ) $ )

Question:
I want to somehow use Riemann-Lebesgue on $ \widehat{f}^{\prime}(n) $ and that'll allow me to finish the attempted proof above; but how? ( how can I show that $ f^{'} \in R(\mathbb{T}) $ ? it is not given that $ f^{'}(0)=f^{'}(2 \pi) $ ), in anycase do you have any idea how to continue to prove the theorem?

Thanks in advance for help!

hazelnut_116
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  • An idea... to be worked out in detail. Replace $f$ by $h=f - g$ where $g$ is a polynomial of low degree such that $h(0)=h(2\pi)$ and $h^\prime(0)=h^\prime(2\pi)$ – mathcounterexamples.net Feb 12 '22 at 10:46
  • If I take $ g $ s.t. $ \hat{g}' = in \cdot f(2\pi) + \frac{1}{2\pi}( f(2\pi) - f(0) ) $ , $ \hat{h}' = \hat{f}' - \hat{g}' = in \cdot f(n) - in \cdot f(2\pi) $
    $ \hat{h}'(0) = \hat{h}'(2\pi) $,so $n\hat{h}(n)-\frac{i}{2 \pi}[h(2 \pi)-h(0)] \underset{n \rightarrow \pm \infty}{\longrightarrow} 0 $ which implies
    $ \lim_{n\to \pm \infty}{n\hat{f}(n)-\frac{i}{2\pi}[f(2 \pi)-f(0)]} = \lim_{n\to\pm \infty}{ \frac{1}{i}\cdot\widehat{f}^{\prime}(n) } = \lim_{n\to \pm \infty}{n \hat{g}(n)-\frac{i}{2\pi}[g(2\pi)-g(0)] }=\lim_{n\to\pm \infty}{\frac{1}{i}\cdot \widehat{g}^{\prime}(n)}$     – hazelnut_116 Feb 12 '22 at 11:10
  • $ \implies \lim_{n\to \pm \infty}{ \widehat{f}^{\prime}(n) } = \lim_{n\to \pm \infty}{ \widehat{g}^{\prime}(n) } $
    But how do I continue to show $ \lim_{n\to \pm \infty}{ \widehat{f}^{\prime}(n) } = 0 $ ? ( Maybe I should've chosen a different $g $ )     – hazelnut_116 Feb 12 '22 at 11:10

1 Answers1

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You are confusing $\hat{f}'$ and $\widehat{f'}$. These are not equal. Your last term is proportional to $\widehat{f'}(n)$. You can apply the Riemann-Lebesgue lemma to $f'$.

coudy
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  • $n \hat{f}(n)-\frac{i}{2 \pi}[f(2 \pi)-f(0)] $ is proportional to $ \widehat{f}^{\prime}(n)$, but why can I apply Riemann-Lebesgue lemma to $f'$? how do I know that $ f'(0) = f'( 2 \pi) $ ? – hazelnut_116 Feb 12 '22 at 10:49
  • The Riemann-Lebesgue holds even if the value at the two endpoints are different. You have to review the proof to check this. The Riemann-Lebesgue actually holds for Riemann-integrable functions defined on the interval $[0, 2\pi]$. – coudy Feb 12 '22 at 10:52