# Cosine series as a Fourier series.

Theorem: Let $a_{n}\downarrow 0$ and suppose that $\left(a_{n}\right)$ is a quasi-convex. Then $\displaystyle \frac{a_{0}}{2}+\sum a_{n}\cos nx$ is the Fourier series of the $L^{1}$ function $\displaystyle f(x)=\sum_{n=0}^{\infty}(n+1)\Delta ^{2}a_{n}\frac{1}{2} K_{n}(x), x\not \equiv 0 \pmod {2\pi}$. Further, $S_{N}(x)\to f(x)$ except at $x\equiv 0 \pmod {2\pi}$, where $\displaystyle S_{N}(x)=\frac{a_{0}}{2}+\sum_{n=1}^{N}a_{n}\cos nx$.

Proof: First we see that if $x \not\equiv 0\pmod {2\pi}$ then $\displaystyle \sum_{n=0}^{\infty}(n+1)\Delta ^{2}a_{n}\frac{1}{2} K_{n}(x) < +\infty$.

Indeed, \begin{equation*} \begin{split} \left|(n+1)\Delta ^{2}a_{n}K_{n}(x_{0})\right| &=\left|n+1\right|\left|\Delta ^{2}a_{n}\right|\left|K_{n}(x_{0})\right|\\ &\leq \left|n+1\right|\left| \Delta ^{2}a_{n}\right|\text{cosec}^{2}\frac {x_{0}}{2} \\ &<+\infty \left(\because \sum_{n=0}^{\infty}(n+1)\left|\Delta^{2}a_{n}\right|<+\infty \right) \end{split} \end{equation*} $\therefore f(x)$ is well-defined everywhere except at $x\equiv 0\pmod {2\pi}.$

We now show that $f\in L^{1}$.

Let $\displaystyle t_{N}(x) =\sum_{n=0}^{N}(n+1)\Delta ^{2}a_{n}\frac{1}{2}K_{n}(x).$

Now, $t_{N}(x)\rightarrow f(x)$ a.e.

$\therefore \left|t_{N}(x)\right| \rightarrow \left|f(x)\right|$ a.e.

By Fatou's lemma, \begin{equation*} \begin{split} \therefore \frac{1}{\pi} \int_{-\pi}^{\pi}\left|f(x)\right|dx &\leq \frac{1}{\pi}\lim_{N\to\infty} inf \int_{-\pi}^{\pi}\left|t_{N}(x)\right|dx\\ &\leq \frac{1}{\pi}\lim_{N\to\infty}inf\int_{-\pi}^{\pi}\frac{1}{2}\sum_{n=0}^{N}(n+1)\left|\Delta ^{2}a_{n}\right|\left|K_{N}(x)\right|dx\\ &=\frac{1}{2} \lim_{N\to\infty}inf \left[\sum_{n=0}^{N}(n+1)\left|\Delta ^{2}a_{n}\right|\right]\\ &=\frac{1}{2} \sum_{n=0}^{\infty}(n+1)\left|\Delta^{2}a_{n}\right| \\ &<+\infty . \end{split} \end{equation*} $\therefore f\in L^{1}$.

We now show that $\|t_{N}-f \|_{1} \rightarrow0$ as $N \to \infty$. \begin{equation*} \begin{split} \left|t_{N}(x)-f(x) \right| &=\left|\sum_{n=N+1}^{\infty} (n+1)\Delta^{2}a_{n}\frac{1}{2}K_{n}(x)\right|\\ &\leq \sum_{n=N+1}^{\infty}\left|(n+1)\Delta^{2}a_{n}\frac{1}{2}K_{n}(x)\right|\\ \therefore \left\|t_{N}-f\right\|_{1} &\leq \left\|\sum_{n=N+1}^{\infty}\left|(n+1)\Delta^{2}a_{n}\frac{1}{2}K_{n}(x)\right| \right\|_{1}\\ &\leq\sum_{n=N+1}^{\infty}(n+1)\left|\Delta^{2}a_{n} \right|\frac{1}{2} \rightarrow 0\;{\text {as}}\; N\to\infty.\ (\because \|K_{n}\|_{1}=1.) \end{split} \end{equation*} $\therefore \left\|t_{N}-f\right\|_{1} \rightarrow 0$ as $N\to\infty$.

Next we show that $S_{N}(x)\to f(x),$ for $x\not\equiv 0 \pmod{2\pi}.$

Now, $\displaystyle S_{N}(x)=\frac{a_{0}}{2}+\sum_{n=1}^{N}a_{n}\cos nx.$

$\displaystyle \therefore S_{N}(x) =\sum_{n=0}^{N}a_{n}b_{n}(x)$ where $b_{n}(x)= \begin{cases} \cos nx & ,n\neq0\\ \frac{1}{2} & ,n=0 . \end{cases}$

Applying partial summation formula twice, we get, \begin{equation*} \begin{split} S_{N}(x) &= a_{N}\frac{1}{2}D_{N}(x)+\sum_{n=0}^{N-1}\Delta a_{n}\frac{1}{2} D_{N}(x)\\ &=a_{N}\frac{1}{2}D_{N}(x)+\frac{1}{2}\left[\Delta a_{N-1}N K_{N-1}(x)+\sum_{n=0}^{N-2}\Delta^{2}a_{n} (n+1)K_{n}(x)\right]. \end{split} \end{equation*} Now for fixed $x\not\equiv 0\pmod {2\pi},\ D_{n}(x)\; \text{and} \; K_{n}(x)$ are bounded sequence of numbers.

Since $(a_{n})$ is convex, $N\Delta a_{N}\to 0$ as $N\to \infty$. Also $a_{n}\to 0$.

$\displaystyle \therefore \lim_{N\to\infty} S_{N}(x)=f(x)$.

It remains to show that the series $\displaystyle \frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_{n}\cos nx$ is a Fourier series of $f(x)$.

Here $\displaystyle S_{N}(x)=\frac{a_{0}}{2}+\sum_{n=1}^{N}a_{n}\cos nx.$ \begin{equation*} \begin{split} \therefore \frac{1}{\pi}\int_{-\pi}^{\pi}S_{N}(x)\cos kx dx &= \frac{1}{\pi}\int_{-\pi}^{\pi}\left[\frac{a_{0}}{2}+\sum_{n=1}^{N}a_{n}\cos nx\right]\cos kx dx.\\ &=a_{k}. \end{split} \end{equation*} But $S_{N}(x)\to f(x)$ as $N\to\infty$.

$\displaystyle \therefore a_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos kx dx.$

Now, $\displaystyle f(x)=\sum_{n=0}^{\infty}(n+1) \Delta^{2}a_{n}\frac{1}{2}K_{n}(x),$ which is an even function. ($\because K_{n}(x)$ is an even function.)

$\therefore b_{k}=0.$

Am i right? If i am doing any mistake then please suggest me! Thanks in advance!

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You should perhaps explain some of your symbols: what's $\,\Delta^2\,$? Or maybe it was meant to be $\,\Delta_{a_n}\,$? What is $\,K_n\,$? Sombody's Kernel (Fejer's, Dirichlet's...)? – DonAntonio Jun 30 '12 at 11:53
Ok! $$\Delta^{2}a_{n}=\Delta a_{n}-\Delta a_{n+1}.$$ $$K_{n}(x)= \frac{1}{n+1}\sum_{k=0}^{n} D_{k}(x),$$ which is known as nth Fejer's kernel. And $a_{n}\downarrow 0$ means $a_{n}$ is decreasing and converges to $0$. – Kns Jul 1 '12 at 6:11
Kns: What happened there? – Did Jul 1 '12 at 10:26