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A trigonometric series $ \displaystyle \frac{a_0}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx +b_{n}\sin nx)$ is a Fourier series of a essentially bounded function if and only if there exists a constant $K$ such that $|\sigma_{n}(x)|\leq K$ almost everywhere.

Proof:

Let us assume that a trigonometric series $ \displaystyle \frac{a_0}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx +b_{n}\sin nx)$ is a Fourier series of a essentially bounded function $f$.

Since $f$ is essentially bounded, \begin{equation} |f(x)|\leq M\ \text{almost everywhere}.\tag{1} \end{equation} Now we know that \begin{equation} \sigma_{n}(x) =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x+u)K_{n}(u)du,\tag{2} \end{equation} where $K_{n}(u)$ is Fejer's kernel.

By $(1)$ and $(2)$, $\displaystyle |\sigma_{n}(x)|\leq M$ almost everywhere. $(\because \displaystyle \frac{1}{\pi}\int_{-\pi}^{\pi}K_{n}(t)dt=1$.)

Conversely let us assume that Cesaro means $\sigma_{n}(x)$ of a trigonometric series $\displaystyle \frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)$ is esssentialy bounded.

i.e. $|\sigma_{n}(x)|\leq K$ almost everywhere.

$\displaystyle \therefore \frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x)dx \leq 2K^{2}$.

We know that,\ $\displaystyle \sigma_{n}(x)=\sum_{k=0}^{n}\left(1-\frac{k}{n+1}\right)(a_{k}\cos kx +b_{k}\sin kx)$.

By Parseval's identity,

\begin{equation*} \frac{1}{\pi}\int_{-\pi}^{\pi}\sigma_{n}^{2}(x) dx =\frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left( 1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2}). \end{equation*} \begin{equation*}\therefore \frac{a_{0}^{2}}{2}+\sum_{k=1}^{n}\left( 1-\frac{k}{n+1}\right)^{2}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}. \end{equation*} Taking $n\to\infty$, we get \begin{equation*} \frac{a_{0}^{2}}{2}+\sum_{k=1}^{\infty}(a_{k}^{2}+b_{k}^{2})\leq 2K^{2}. \end{equation*} Therefore the given trigonometric series under consideration is a Fourier series of some function $f(x)\in L^{2}$. ($\because$ Riesz-Fischer theorem)

But since $\sigma_{n}(x)\to f(x)$ almost everywhere and $|\sigma_{n}(x)|\leq K$, it follows that $|f(x)|\leq K$ almost everywhere.

i.e. $f(x)$ is essentially bounded.

Am i rightly approach the proof of this theorem? Your's suggestion is very helpful to me.

share|improve this question
    
1) Uniform convergence is not the same as uniform boundedness. 2) you can't so easy take a limit $n\to \infty$. The reason is that $n$ is contained in summed terms. –  Norbert Jun 18 '12 at 6:28
    
Also, What is $m$? You defined $M$ but not $m$. –  Willie Wong Jun 18 '12 at 8:00
    
I edit this theorem statement and and i change some parts of this proof of theorem to get meaningful things! Am i right? –  Kns Jun 18 '12 at 15:28

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