How to show that if all fourier coefficient of a function is zero, then the function is zero function?

Question

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?

Answer 1

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 $x$.

Answer 2

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 $f\equiv 0.$

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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 $f\equiv 0.$

Answer 3

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.