Given the Bessel inequality, $$\sum_{n=1}^N (c_{n})^2\leq ||f||^{2}{ }$$ $c_{n}=(f,\phi_n)$ where {$\phi_n$} is an orthonormal set in $C_{p}(0,\pi)$
$||f||^2=(f,f)$ an inner product space
(a) Obtain the following inequality
$$\frac{a_{0}^2}{2}+\sum_{n=1}^N (a_{n})^2\leq \frac{2}{\pi}\int_0^\pi [f(x)]^2 dx\ $$
(b) Also show how it follows from the bessel inequality that $$\sum_{n=1}^N (b_{n})^2\leq \frac{2}{\pi}\int_0^\pi [f(x)]^2 dx\ $$
This is what i tried
First $||f||$ being an an inner product equals to $$\frac{2}{\pi}\int_0^\pi [f(x)]^2!dx\ $$
Next we let $a_{n}$,$b_{n}$ be the fourier coefficients of the fourier cosine and sine series representation corresponding to $F\in C{p(0,\pi)}$
We have $c_{n}=(f,\phi_n)$ where {$\phi_n$} is an orthonormal set in $C_{p}(0,\pi)$
$a_{n}=\frac{2}{\pi}\int_0^\pi [f(x)]\cos(nx)dx=\frac{2}{\pi}(f,\phi_{n})=\frac{2}{\pi}c_{n} $
$b_{n}=\frac{2}{\pi}\int_0^\pi [f(x)]\sin(nx)dx$
while $a_{0}=\frac{2}{\pi}\int_0^\pi [f(x)]dx \frac{2}{\sqrt\pi}(f,\phi_{n})=\frac{2}{\sqrt\pi}c_{0} $
Also if $\phi_{0}=\frac{1}{\sqrt{\pi}}$ and $\phi_{n}=\sqrt{\frac{2}{\pi}\ cos (nx)}$, then {$\phi_{n}$} is an orthonormal set
Now what remains is how do i piece all the info above to solve the question. Would anyone expalin .Thanks
There are shorter ways to do this, but I'll do the computations so there is less confusion. This is essentially deriving the Bessel inequality, but from this approach you should hopefully you can answer your questions. I realise this is probably overkill for this problem, but whatever. We have our Fourier series given by
$$f = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos(nx) + b_{n} \sin(nx)$$
and define
$$f_{N} = \frac{a_{0}}{2} + \sum_{n = 1}^{N} a_{n} \cos(nx) + b_{n} \sin(nx)$$
Looking at the difference, we have
\begin{align} \| f - f_{N} \|^{2} &= \int_{- \pi}^{\pi} (f - f_{N})^{2} dx \\ &= \int_{- \pi}^{\pi} (f^{2} - 2 f f_{n} + f_{N}^{2}) dx \ge 0 \quad (*) \end{align}
Now, we can explicitly compute the integrals of $f_{N}^{2}$ and $f f_{N}$ (I'll do the $f_{N}^{2}$ case, you can do the other). We have
\begin{align} &\int_{- \pi}^{\pi} f_{N}^{2} dx \\ &= \int_{- \pi}^{\pi} \left ( \frac{a_{0}}{2} + \sum_{n = 1}^{N} a_{n} \cos(nx) + b_{n} \sin(nx) \right )^{2} dx \\ &= \underbrace{\int_{- \pi}^{\pi} \left( \frac{a_{0}}{2} \right)^{2} dx}_{I_{1}} + \underbrace{\int_{- \pi}^{\pi} a_{0} \sum_{n = 1}^{N} a_{n} \cos(nx) + b_{n} \sin(nx) dx}_{I_{2}} + \underbrace{\int_{- \pi}^{\pi} \left ( \sum_{n = 1}^{N} a_{n} \cos(nx) + b_{n} \sin(nx) \right )^{2} dx}_{I_{3}} \end{align}
Now, the first integral $I_{1} = \pi a_{0}^{2}/2$. Interchanging the sum and integral in $I_{2}$ (why can we do this?) and noting that $a_{n}, b_{n}$ are constants, we can compute the integrals of $\cos(nx)$ and $\sin(nx)$ over $(- \pi, \pi)$, which both evaluate to zero. Hence, $I_{2} = 0$. For $I_{3}$, we interchange the sum and integrals again and note that by orthogonality, all the products of $\sin$ and $\cos$ evaluate to zero. Hence, the only integrals that are nonzero are
\begin{align} \int_{- \pi}^{\pi} a_{n}^{2} \cos^{2}(nx) dx &= \pi a_{n}^{2} \\ \int_{- \pi}^{\pi} b_{n}^{2} \sin^{2}(nx) dx &= \pi b_{n}^{2} \\ \end{align}
and so
\begin{align} \int_{- \pi}^{\pi} f_{N}^{2} dx &= I_{1} + I_{3} \\ &= \pi \left( \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \right) \\ \end{align}
Similarly,
$$ \int_{- \pi}^{\pi} f f_{N} dx = \pi \left( \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \right)$$
and so $(*)$ becomes
\begin{align} \int_{- \pi}^{\pi} f^{2} dx - \pi \left( \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \right) &\ge 0 \\ \implies \int_{- \pi}^{\pi} f^{2} dx &\ge \pi \left( \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \right) \\ \implies \frac{1}{\pi} \int_{- \pi}^{\pi} f^{2} dx &\ge \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \end{align}
Now note that $f^{2}$ is even, and hence
\begin{align} \frac{1}{\pi} \int_{- \pi}^{\pi} f^{2} dx = \frac{2}{\pi} \int_{0}^{\pi} f^{2} dx &\ge \frac{a_{0}^{2}}{2} + \sum_{n = 1}^{N} a_{n}^{2} + b_{n}^{2} \end{align}
From this final result, it should be clear how to obtain your inequalities.