For the function $f \in L^2 ([-\pi,\pi])$ define the map $ T: L^2([-\pi, \pi]) \to R $ as $T(f)=a_1+b_1 $ if the Fourier series of $f$ is of the form
$f $ ~ $a_0 +\sum_{n=1}^{\infty} (a_ncos(nx)+b_nsin(nx)) $ find function $h \in L^2 ([-\pi,\pi])$ so that $T(f)=\int_{-\pi}^\pi f(x)h(x)dx $
In order to find $h$ I started the following $ \int_{-\pi}^\pi f(x)h(x)dx=a_1+b_1$ then
$f(x)$~$a_0+a_1 cosx+b_1sinx + \sum_{n=2}^{\infty}(a_n cos(nx)+b_nsin(nx))h(x)dx$
$\int_{-\pi}^\pi a_0h(x)dx=0 $
$ \int_{-\pi}^\pi(\sum_{n=2}^{\infty}(a_ncosnx+b_nsinnx)h(x)dx=0 $
then $ \int_{-\pi}^\pi (a_1cosx+b_1sinx)h(x)dx=a_1+b_1 $
and then i got following: $ \int_{-\pi}^\pi a_1cosxh(x)dx=1$ which will be $\int_{-\pi}^\pi cosxh(x)dx=1$
and same logic for be I got integral sinxh(x) is equal to 1. However, I do not know how to proceed forward is in order to find h(x). Can you please help me out? I could not find intuitive way of solving this problem.
The Fourier coefficients are defined by $$a_0:=\frac1{2\pi}\int_{-\pi}^\pi f(x)\,\mathrm d x\\ a_n:=\frac1{\pi }\int_{-\pi }^\pi f(x)\cos(nx)\,\mathrm d x\\ b_n:=\frac1{\pi }\int_{-\pi }^\pi f(x)\sin(nx)\,\mathrm d x $$ for $n\in \Bbb N_{> 0} $. Therefore $h(x)=\frac1{\pi }(\cos (x)+\sin(x))$.