# If a trigonometric series converges uniformly, to prove its coefficients are the Fourier coefficients

I am new to Fourier series and here is what I understand (please correct me where I am wrong) :

The trigonometric series $a_0/2 + Σ(a_ncos(nx)+b_nsin(nx))$ will be called a Fourier series of $f$ if $a_n$ and $b_n$ are the Fourier coefficients of $f$. It may or may not converge or even if it does the sum may be different from $f$.

I want to show that if the trigonometric series does uniformly converge on $[-π,π]$ to $f$, then the trigonometric series is the Fourier series of $f$. My book does this by multiplying with $cos(nx)$ then integrating the new series term by term and then multiplying with $sin(nx)$ and then does the same. But how do we know that the new series will also be uniformly convergent? If it's not, then there is no reason for us to believe that we can integrate the series term by term.

• What kind of function is $f?$ Also, if $f_n$ converges uniformly and $g$ is bounded, then $f_ng$ converges uniformly. – zhw. Dec 1 '17 at 1:00
• $f$ is the uniform sum function of the trigonometric series in the interval $[-π,π]$. – Hrit Roy Dec 1 '17 at 1:03
• no, that makes no sense. read your own second paragraph. – zhw. Dec 1 '17 at 1:05
• 2nd paragraph is the definition of a Fourier series. It has nothing to do with the next one. – Hrit Roy Dec 1 '17 at 1:06
• I need to show if $f$ is the uniform limit of the trigonometric series in $[-π,π]$, then the trigonometric series (not saying it's a Fourier series yet) is the Fourier series of $f$ i.e., $a_n$ and $b_n$ are the Fourier coefficients of $f$. – Hrit Roy Dec 1 '17 at 1:08

Suppose the partial sums $S_N$ of the trigonometric series converge uniformly on $[-\pi,\pi]$ to $f$:

$$\tag{*}f(x) = \lim_{N \to \infty}S_N(x) =\frac{a_0}{2} + \sum_{k=1}^\infty (a_k \cos kx + b_k \sin kx).$$

By uniform convergence, since $S_N$ is continuous, the limit function $f$ is continuous and we can integrate the series termwise (after multiplying both sides of (*) by $\sin nx$ or $\cos nx$), to obtain

$$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \,dx , \\ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \,dx.$$

Obtaining this result uses the orthogonality of the trigonometric functions.

Thus, $a_n$ and $b_n$ are the Fourier coefficients.

Uniform convergence is sufficient here and, without uniform convergence, not every pointwise convergent trigonometric sum is a Fourier series.

For example, $$\sum_{n=2}^\infty \frac{ \sin nx}{\log n}$$

converges by the Dirichlet test but is not a Fourier series.

• That is exactly where I fail to understand. How can we integrate it termwise after multiplication by $cosnx$? How do we know that the uniform convergence is preserved? – Hrit Roy Dec 1 '17 at 2:12
• A basic theorem of analysis states if $f_n$ is integrable and $\sum f_n$ converges uniformly on $[a,b]$ to $f$, then $f$ is integrable and $\int_a^b \sum_{n=1}^\infty f_n(x) \, dx = \sum_{n=1}^\infty \int_a^b f_n(x) \, dx$. Multiplying by any continuous, bounded function preserves the integrability and the uniform convergence. – RRL Dec 1 '17 at 2:20
• The first part I know. But where are we getting the second part from? The multiplication part? – Hrit Roy Dec 1 '17 at 2:22
• $S_n$ converges uniformly to $f$ if for $\epsilon > 0$ we have $|S_m(x) - S_n(x)| < \epsilon$ for all $x$ in the domain and $m >n$ sufficiently large. Multiplying by bounded $g$ where $|g(x)| \leqslant B$ gives us $|g(x) S_m(x) - g(x) S_n(x)| \leqslant B|S_m(x) - S_n(x)| < B\epsilon$ and we are good. – RRL Dec 1 '17 at 2:32
• It so happens that $\cos$ and $\sin$ are continuous. All we need is that $g$ is bounded for the uniform convergence. Continuity makes everything simpler in the proof. We want $f_n g$ to be integrable so this is sufficient. – RRL Dec 1 '17 at 2:58