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Suppose $f \in L_1(-\pi,\pi)$, satisfying $\hat{f}(n) = 0, \forall n\in \mathbb{Z}$ (which means all Fourier coefficients are $0$). Does $f = 0$ almost everywhere in $(-\pi,\pi)$?

The Fourier coefficients of $f$ are defined by $\hat{f}(n) = \int_{-\pi}^{\pi}f(\theta)e^{-in\theta}d\theta$

This is a similar question, but it does not provide a clear answer: Functions whose Fourier coefficients are all zero

U2647
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  • The Cesaro means of the partial sums of the Fourier series of an $L^1$ function converge to that function in the $L^1$ norm. – Angina Seng Mar 10 '19 at 08:54
  • @LordSharktheUnknown You are right. Apply Fejer's theorem, then your comment is reached. Something in real analysis tells that convergence in $L_p \quad (p \ge 1)$ implies two functions equal a.e.. – U2647 Mar 10 '19 at 12:19
  • A very elementary proof of this is given in Fourier Series by Edwards. – Kavi Rama Murthy Mar 10 '19 at 12:28

2 Answers2

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This follows immediately from Parseval's identity, $$ \int_{-\pi}^\pi |f(\theta)|^2d\theta = \sum_{n=0}^\infty |\hat{f}(n)|^2. $$ Since the coefficients are zero, so is the function's square integral, and thus so is the function.

eyeballfrog
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    Doesn't that require the function to be $L^2$? – Angina Seng Mar 10 '19 at 08:38
  • @LordSharktheUnknown We can say $|f(\theta)|^2 d \theta$ is a measure, $|.|{L^2}$ is a well-defined map $L^1 \to \mathbb{R}{\ge 0} \cup \infty$ and Parseval stays true, for example by looking at $\lim_{A \to \infty} f 1_{|f| \le A}$ which converges to $f$ both in $L^1$ and $L^2$ (when $f \in L^2$) – reuns Mar 10 '19 at 09:07
  • @reuns I'm sorry for digging a very old answer of yours but I got a question to ask, if you don't mind. As a matter of fact, I don't understand your argument of how letting $A\to\infty$ works for the term on the RHS, i.e. $\sum_{n=0}^\infty |\hat{f}(n)|^2$. The LHS converges nicely by MCT but wouldn't the Fourier coefficients on the RHS keep changing for each different $A$ in a non-monotonic fashion? How exactly did you plan to deal with this? – BigbearZzz Jul 07 '19 at 00:59
  • @reuns Yes, we know that $\sum_n |\hat{f 1_{|f|< A}}(n)|^2 \to \infty$, but how do you relate that to the Fourier coefficients of $f$? – BigbearZzz Jul 07 '19 at 01:28
  • @reuns I think I see what you meant now, we can just use DCT to get $\hat{f 1_{|f|< A}}(n) \to \hat f(n)$. Is this what you had in mind? I forgot that $\sin, \cos$ are bounded function and was thinking of them in term of general $L^2$ functions. – BigbearZzz Jul 07 '19 at 01:37
  • @BigbearZzz Sorry I'm lost. We have a periodic distribution $f$ whose Fourier coefficients are $c_n(f)$ of at most polynomial growth, then $f$ is $L^2$ iff $\sum_n |c_n(f)|^2 < \infty$, the proof is that $f \ast \phi_\epsilon \to f$ where say $c_n(\phi_\epsilon \ast f ) = e^{-n^2 \epsilon^2} c_n(f)$. Iff $\sum_n |c_n(f)|^2 = \infty$ we can set $|f|{L^2}^2 = \infty$ it is compatible with $\lim{\epsilon \to 0} | f \ast \phi_\epsilon|2$ and $\lim{A \to \infty} |f 1_{|f| < A}|_2$, because finiteness of one implies (Parseval) finiteness of the others – reuns Jul 07 '19 at 01:40
  • @BigbearZzz So yes : if $\sum_n |c_n(f)|^2 < \infty$ then $\lim_{A \to \infty}\sum_n |c_n(f 1_{|f|< A})| ^2 $ converges by DCT, if $\sum_n |c_n(f)|^2 = \infty$ then $|f|2$ can't be finite so $\lim{A \to \infty} \sum_n |c_n(f 1_{|f|< A})| ^2 =\lim_{A \to \infty}|f 1_{|f|< A}|_2^2 =|f|_2^2= \infty$ – reuns Jul 07 '19 at 01:47
  • @reuns I am quite interested in your method. The way I'd prove the OP's problem would be to invoke the Riesz-Markov theorem to show that the measure induced by $f$ is zero, using the density of trigonometric polynomials in $C(\Bbb T)$. That would work for a general signed Borel measure but it seems like you're saying that the result is even true for perhaps more general distributions as well. Is there a place where I can read about such general results? – BigbearZzz Jul 07 '19 at 02:00
  • @reuns And may I also ask why the "polynomial growth" assumption is needed if we assume $f$ to be a distribution? – BigbearZzz Jul 07 '19 at 08:53
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    To apply Parseval you need $f \in L^2$. This is not an answer. – copper.hat Jul 24 '21 at 15:17
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We know that a $L^1(-\pi, \pi)$ function's continuous point is dense in $(-\pi,\pi)$.

Use the Theorem 2.1 in Stein's 'Fourier Analysis'(Page 39), we know that when $\hat{f}(n)=0$ for all $n\in \mathbb{Z}$, $f(x)=0$ at all continuous point, and hence $f=0$ a.e. $x\in (-\pi,\pi)$.

IMOS
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