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Let $f$ be a continuous and integrable function with period $2\pi$. Consider its fourier coefficients with respect to the orthonormal system $\{ \frac {1}{\sqrt{2\pi} } e^{inx}\}$. If all the Fourier coefficients are zero, prove that $f$ is the zero function.

I think it is a very natural proposition but I find myself stuck because we cannot say that $f$ is equal to its Fourier series. Are there any simple and fast way to prove this? Or this problem is harder than it seems?

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  • @copper.hat What do you mean by “almost"? –  May 03 '15 at 22:43
  • Suppose you omit just one value of $n$, say $n=6$, from the list of functions $\frac 1 {\sqrt{2\pi}} e^{inx}$. Then the proposition becomes false: there is a continuous function whose coefficients with respect to the orthonormal system are all $0$, but which is not the zero function. Namely, the function $x\mapsto e^{i6x}$. So somehow you need to show that no necessary function is omitted from this list of orthonormal functions. And probably it is harder than it looks. ${}\qquad{}$ – Michael Hardy May 03 '15 at 22:54
  • "integrable" follows from continuity, so its inclusion in the hypotheses is redundant. – zhw. May 03 '15 at 22:56
  • Integrability follows from continuity in this case because the domain $\mathbb Z/2\pi$ is compact. But lets not leave an impression that integrability follows from continuity in all instances. ${}\qquad{}$ – Michael Hardy May 03 '15 at 22:58

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Let $$S_n(x) = \sum_{m=-n}^{n}\hat {f} (m)[\frac {1}{\sqrt{2\pi}}e^{imx}].$$

Then the Fejer sums $\frac{1}{2N+1}\sum_{n=-N}^{N}S_n(x)$ converge uniformly to $f$ on $[0,2\pi].$ If all $\hat {f}(m)=0,$ then it follows that $f\equiv 0.$

Added later: Here's another proof: On the unit circle $\mathbb {T},$ the algebra of trigonometric polynomials $\sum_{n=-N}^{N}a_n\zeta ^n$ is, by Stone-Weierstrass, dense in $C(\mathbb {T}).$ If $\hat {f}(n) = 0$ for all $n,$ then it's easy to see $\int_{-\pi}^\pi fp = 0$ for every trig. polynomial $p.$ Choose a sequence of trig polys $p_N \to \bar {f}$ uniformly on $\mathbb {T}$ to see $\int_{-\pi}^\pi |f|^2 = 0,$ proving $f\equiv 0.$

zhw.
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  • Why do you sum twice over the series? Also, if the Fourier series converges uniformly, can we say that it converges to f? I think that is the key to prove this but I cannot find why it is true. –  May 03 '15 at 23:10
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    The Fourier series of $f$ need not even converge to $f$ pointwise everywhere. So forget about uniform convergence of the FS. The Fejer means of the partial sums of the FS do converge uniformly to $f.$ – zhw. May 03 '15 at 23:18
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    @RSBST: zhw is using the fact that the Fejer kernel gives rise to an approximate identity. It is a general result that the convolution of an approximate identity with a continuous function $f$ converges uniformly to $f$ on compact sets. Since $\mathbb{T}$ is compact, the convolution converges uniformly on $\mathbb{T}$. – Matt Rosenzweig May 03 '15 at 23:26
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    I added a Stone-Weierstrass approach to my original answer. – zhw. May 03 '15 at 23:39
  • Small question: Do we need uniform convergence of the Fejer sums to $f$? We know it's true but pointwise convergence of the Fejer sums would suffice for the purpose of the proof right? – John11 May 02 '18 at 21:47
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    @John11 Yes,it would, but I don't see how you get a pointwise result without getting the uniform result for free. – zhw. May 02 '18 at 23:08
  • @zhw True. Thank you! – John11 May 03 '18 at 10:11
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Let $e_n(x) = \frac {1}{\sqrt{2\pi} }e^{inx}$, and let $E=\{e_n\}_{n \in \mathbb{Z}}$.

The answer here shows that the system $E$ is dense in $L_2[0, 2 \pi]$.

Suppose $f$ is such that $f \bot e_n$ for all $n$, then we have $f \in (\overline{\operatorname{sp}} E)^\bot$, which is $\{0\} \subset L_2[0,2 \pi]$, since $E$ is dense.

Then $f = 0$ in $L_2[0, 2 \pi]$, which means $f(x) = 0$ ae. $x \in [0,2 \pi]$.

Since $f$ is continuous, we have $f(x) = 0$ for all $x$.

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copper.hat
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  • If the inner product of f with a complete system is zero, then f is zero(over the interval) , then f is zero? Sorry that I do not understand what does it mean by complete system. Can you give me a link of the theorem? –  May 03 '15 at 22:55
  • This is just an assertion. The thing to be proved is that the system is complete. ${}\qquad{}$ – Michael Hardy May 03 '15 at 22:55
  • @RSBST: Your question is tantamount to showing that the $e_n$ are complete, so I haven't really answered you question. If you combine Bessel's inequality with the Riemann Lebesgue lemma then you can obtain the result. Let me see if I can find a simpler proof. – copper.hat May 03 '15 at 22:58
  • I think the Bessel inequality somehow gives a inequality of an opposite direction( we knot all coefficients are zero and the sum of their squares are less then the integral of the square of f) I think I am a little bit lost. –  May 03 '15 at 23:06
  • @RSBST: Sorry, sloppy thinking on my part. I have changed my answer. Another useful (but more difficult to prove) result is Carleson's theorem. – copper.hat May 03 '15 at 23:34
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One can appeal to the completeness of the system given, a consequence is that f is equal to the zero function almost everywhere, i.e., it is zero almost everywhere. But then the only continuous null function is the zero function. (See e.g. Hardy and Rogosinski Fourier series, section 1.11 to 2.6 page 13-19, particularly Theorem 8 and Theorem 19 or for the proof of the completeness.)

One may also note that since all the Fourier coefficients are zero, the Fourier series converges boundedly to the zero function, consequently it is the Fourier series of its sum function, the zero function (obvious). Since f is continuous, by Fejer's theorem its Fourier series is (C,1) summable to f(x) uniformly. Plainly, the (C,1) means are all zero, so the same Fourier series is also (C,1) summable to the zero function everywhere. Hence f is identically zero.