Show that if $f$ and $g$ have the same Fourier series, and f, g are both continuous, then f = g on $[-\pi, \pi]$. (f, g has periodic $2\pi$)

Question

By using a suitable convergence theorem, show that if $f$ and $g$ have the same Fourier series, and f, g are both continuous, then f = g on $[-\pi, \pi]$. (f, g has periodic $2\pi$)

If the partial sum of the Fourier Series, say, $S_n(f;x)$, which is (by assumption) equal to $S_n(g;x)$, converges uniformly to function $f$ and $g$, then the problem would be easily solved:

given $\epsilon\gt0, \exists N_1,N_2\in \mathbb{N}$, such that when$n\gt N_1, N_2$, respectively we have

$|f-S_n(f;x)| \le {\epsilon\over2} , |g - S_n(f;x)| \le {\epsilon\over2} $

then, $|f - g| \le |f-S_n(f;x)| + |g - S_n(f;x)| \le {\epsilon\over2} + {\epsilon\over2} = \epsilon$ for $\forall n\gt max\{N_1, N_2\}$

However, I am aware that the above argument is based on the belief that the partial sum will uniformly converge. To prove the uniformity, I find this theorem:

Theorem: The Fourier series of a 2π-periodic continuous and piecewise smooth function converges uniformly.

But apparently, we lack the information to verify whether the conditions will meet. So what should I do?

Answer 1

You cannot use uniform convergence. But $S_n(f)$ converges to $f$ as well as $g$ in $L^{2}([-\pi,\pi])$ and this implies $f=g$ almost everywhere. By continuity $f(x)=g(x)$ for all $x$.

Answer 2

If $f$ and $g$ have the same Fourier series then $$\int^{\pi}_{-\pi}p(t)(f(t)-g(t))\,dt=0$$ for any trigonometric polynomial $p$, for $c_n(f)=\frac{1}{2\pi}\int^\pi e^{-inx}f(x)\,dx =\frac{1}{2\pi}\int^\pi e^{-inx}g(x)\,dx =c_n(g)$ for all $n\in\mathbb{Z}$). By Stone-Weierstrass theorem (complex version) $$\int^{\pi}_{-\pi}\varphi(t)(f(t)-g(t))\,dt=0$$ for any continuous function $\phi$. In particular, when $f$ and $g$ are continuous, for $\phi(t)=\overline{f(t)-g(t)}$ $$\int^{\pi}_{-\pi}|f(t)-g(t)|^2\,dt=0$$ Hence $f(t)=g(t)$ almost surely and so, $f\equiv g$ (by continuity).