$f$ is continuous on the circle, and its Fourier series is absolutely convergent. Then its Fourier series converges uniformly to $f$.

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Corollary 2.3 in Fourier Analysis: An Introduction, Stein and Shakarchi, is not sufficiently detailed for me to fully understand.

I can see why the Fourier series uniformly converges to some continuous $g$ but how can I justify the equality between $\hat{g}(n)$ with $\hat{f}(n)$? (its Fourier coefficients).

I am not looking for hints, I tried to solve it myself following the "sketch" depicted in the book with no success. Thank you in advance for any help.

the "sketch" depicted in the book. "the Fourier coefficients of g are precisely $f_{n}$ since we can interchange the infinite sum with the integral(a consequence of the uniform convergence of the series)."

Update : currently stuck here \begin{equation} \hat{g}(n) = \frac{1}{2\pi}\int_0^{2 \pi} g(x)\cdot e^{-inx}dx \\ = \frac{1}{2\pi}\int_0^{2 \pi} (\sum_{m \in Z} \hat{f}(m)\cdot e^{imx})\cdot e^{-inx} dx \end{equation}

trying to get $\hat{g}(n) = \hat{f}(n)$

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Uniform convergence is enough to justify interchange of an integral over a finite interval and an infinite sum. If $\sum \hat{f}(n)e^{inx}$ is absolutely convergent then $\sum |\hat{f}(n)|< \infty$ and so the Fourier series converges uniformly for $0 \leq x \leq 2\pi$. Call the sum $h$. Then $h$ is continuous periodic function. Now multiply by $e^{ikx}$, integrate from $o$ to $2\pi$ and take the integral inside the sum. You will get $\hat{h}(k)=\hat{f}(k)$.