# a trigonometric series uniformly converge to a function

if a trigonometric series uniformly converge to a function, is it the Fourier series of the function? I understand the Uniqueness of Fourier Series, but that one is saying if I have 2 continuous functions and they have the same Fourier Coefficients, then these 2 functions are the same. But is it related to my question?

• I have seen an an example of a series which pointwisely converge at every point but it is not a fourier series of any integrable function. I don't know whether it would happen for the uniformly convergence. – Haiyi Tan Oct 21 '16 at 11:30
• Let $p_n$ be the $n$-th partial sum of the series, and $f$ the limit function. Consider the sequence $\int_0^{2\pi} p_n(t)\cos (kt)\,dt$. – Daniel Fischer Oct 21 '16 at 11:34

Yes.

As Daniel Fischer has suggested in his comment, for simplicity, we consider $[0,1]$ and $f=a_0+\sum_{n=1}^\infty a_n\cos(2\pi n x)+b_n\sin(2\pi n x)$, where the trigonometric series converges uniformly to $f$ on $[0,1]$, which implies $f$ is continuous on $[0,1]$ and hence integrable.

Then the Fourier series of $f$ is given by $$f\sim c_0/2+\sum_{i=1}^\infty c_n\cos(2\pi n x)+d_n\sin(2\pi n x),$$ where $c_n= 2\int_{0}^{1} f(x) \cos(2\pi n x)\ dx$ and $d_n= 2\int_{0}^{1} f(x) \sin(2\pi n x)\ dx$ for $n\ge 0$. Then it follows directly from the uniform convergence of the original trigonometric series that $c_n=a_n, d_n=b_n$ for $n\ge 0$, which proves your claim.