Answer: Vanishing Fourier Coefficients Force Zero

Key Idea / Intuition

The key insight is a two-step bootstrap: trigonometric polynomials are dense in continuous functions (Weierstrass's theorem), so if $f$ is orthogonal to every complex exponential $e^{inx}$, it is orthogonal to every trig polynomial, and hence to itself — which forces $\|f\|^2 = 0$.

More concretely, the Weierstrass approximation theorem for trig polynomials says we can find $p_n \to f$ uniformly. Since $\int f \cdot e^{ikx} = 0$ for all $k$, we get $\int f \cdot p_n = 0$ for all $n$. Passing to the limit gives $\int f \cdot f = 0$.

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Formal Proof / Solution

Step 1: Orthogonality extends to all trig polynomials.

A trigonometric polynomial is any finite linear combination $T(x) = \sum_{|k| \le N} c_k e^{ikx}.$ By linearity of the integral and the hypothesis $\hat{f}(k) = 0$ for all $k$: $\int_0^{2\pi} f(x)\, T(x)\, dx = \sum_{|k|\le N} \bar{c}_k \int_0^{2\pi} f(x)\, e^{ikx}\, dx = \sum_{|k|\le N} \bar{c}_k \cdot 2\pi\, \overline{\hat{f}(-k)} = 0.$

Wait — more cleanly: $\int_0^{2\pi} f(x)\, e^{ikx}\,dx = 2\pi\,\widehat{f}(-k) = 0$, so indeed $\int f \cdot T = 0$ for every trig polynomial $T$.

Step 2: Approximate $\overline{f}$ uniformly by trig polynomials.

By the Weierstrass theorem for trigonometric polynomials (which follows from the ordinary polynomial version via $x \mapsto e^{ix}$, or from Fejér's theorem on Cesàro means of the Fourier series of continuous functions):

For any $\varepsilon > 0$, there exists a trig polynomial $T_\varepsilon$ such that $\|T_\varepsilon - \overline{f}\|_\infty < \varepsilon.$

Step 3: Conclude $f \equiv 0$.

Consider $\int_0^{2\pi} |f(x)|^2\, dx = \int_0^{2\pi} f(x)\, \overline{f(x)}\, dx.$

Write $\overline{f} = T_\varepsilon + (\overline{f} - T_\varepsilon)$. Then: $\int_0^{2\pi} f(x)\,\overline{f(x)}\,dx = \underbrace{\int_0^{2\pi} f(x)\,T_\varepsilon(x)\,dx}_{=\,0 \text{ by Step 1}} + \int_0^{2\pi} f(x)\,(\overline{f(x)} - T_\varepsilon(x))\,dx.$

The error term satisfies: $\left|\int_0^{2\pi} f(x)\,(\overline{f(x)} - T_\varepsilon(x))\,dx\right| \le \|f\|_\infty \cdot \|\overline{f} - T_\varepsilon\|_\infty \cdot 2\pi < \|f\|_\infty \cdot \varepsilon \cdot 2\pi.$

Since $\varepsilon > 0$ is arbitrary: $\int_0^{2\pi} |f(x)|^2\, dx = 0.$

Since $|f|^2$ is continuous and non-negative, this forces $f \equiv 0$. $\blacksquare$

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Why this is the "right" proof: It reveals that the true reason is density, not Hilbert space abstraction. The continuous functions are a metric space; the trig polynomials are a dense subspace; and the linear functional $T \mapsto \int fT$ is bounded and zero on a dense set, hence zero everywhere — including when $T = \overline{f}$.