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Question: If the trigonometric series $\displaystyle\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)$ converges uniformly to $g(x)$ on $[-\pi,\pi]$ then $g$ is continuous on $[-\pi,\pi]$ and moreover the given trigonometric series is a Fourier series of $g(x)$.

Answer: The $n$th partial sum of the trigonometric series $\displaystyle\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)$ is given by, $$S_{n}(x)=\frac{a_{0}}{2}+\sum_{k=1}^{n}(a_{k}\cos kx+b_{k}\sin kx).$$ We are given that $S_{n}(x)$ uniformly converges to $g(x)$ on $[-\pi,\pi]$.

Since each $S_{n}(x)$ is continuous and $S_{n}(x)\to g(x)$ uniformly, $g(x)$ is continuous.

Now, $\displaystyle \int_{-\pi}^{\pi}|g(x)-S_{n}(x)|dx\to 0$ as $n\to\infty.$

$\therefore$ For any $m$, we have, $$\int_{-\pi}^{\pi}(g(x)-S_{n}(x))\cos mx dx\to 0\ \text{as}\ n\to\infty,$$ and $$ \int_{-\pi}^{\pi}(g(x)-S_{n}(x))\sin mxdx\to 0\ \text{as}\ n\to\infty.$$ Thus we have, $$ \lim_{n\to\infty}\int_{-\pi}^{\pi}S_{n}(x)\cos mxdx=\int_{-\pi}^{\pi}g(x)\cos mxdx,$$ and $$\lim_{n\to\infty}\int_{-\pi}^{\pi}S_{n}(x)\sin mxdx=\int_{-\pi}^{\pi}g(x)\sin mxdx.$$ If $n\geq m$, we have, \begin{align*} \int_{-\pi}^{\pi}S_{n}(x)\cos mxdx&= \int_{-\pi}^{\pi}(\frac{a_{0}}{2}+\sum_{k=1}^{n}(a_{k}\cos kx+b_k\sin kx)\cos mx dx\\ &= \pi a_{m}.(\because\ \text{Orthogonality relation}) \end{align*} $\displaystyle \therefore \lim_{n\to\infty}S_{n}(x)\cos mxdx=\pi a_{m}\Rightarrow a_{m}=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos mx dx.$

Similarly, we can get,

$b_{m}=\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin mx dx.$ It follows that, the given trigonometric series is a Fourier series of function $g(x)$.

Is there any mistek in my answer?

Note: This result gives me a sufficient condition for the trigonometric series to be a Fourier series.

share|improve this question
    
In the partial sum, the upper limit of the summation should be $n$ instead of $\infty$. –  joriki Apr 21 '12 at 10:55
    
Yes my mistek i am edit it. What about answer? Is everything right? –  Kns Apr 21 '12 at 11:27
    
I think so. I can't find any error. The word is "mistake", by the way. –  joriki Apr 21 '12 at 13:24

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