Letting $a_n$ denote the coefficients in the Fourier cosine series of any function $f(x)$ on $(0,\pi)$ , how can I show that:
$\sum_{n=1}^N c_n^2 \le ||f∥^2$, where $c_n$ are the Fourier constants of $f(x)$ that:
$ a_0^2 \over 2 $+ $ \sum_{n=1}^Na_n^2 \le $ $ 2\over π $ $\int_{0}^{π}(f(x))^2\,dx $, $N = 1,2,3,4\dots$
now that looks better :)
But...this is only part of Bessel's inequality, isn't it? I mean, we have the (complete) orthonormal trigonometric system (basis) $\;\{u_n\}_{n\in\Bbb N}\;,\;\;u_n=\{1,\sin x,\cos x, \sin 2x,\cos2x,\ldots\}\;$ with which we work in Fourier series, so then denoting by $\;\langle,\rangle\;$ the usual inner product we get
$$\sum_{n=1}^\infty \langle f(x),u_n\rangle^2\le||f||^2$$
But using the usual notation from Fourier series the above is just
$$||f||^2\ge\sum_{n=1}^\infty c_n^2 $$