Is T a continuous linear map?

Question

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)) $

How to prove that T is a continuous linear map?

I took the limit to infinity to show that $lim T(f-f_n) = T(f)-lim T(f_n)=a_1+b_1-(a_1+b_1)=0$ it is what intuition suggests me, however i do not have a ground for such.

Answer 1

Since$$a_1=\frac1\pi\int_{-\pi}^\pi f(t)\cos(t)\,\mathrm dt=\frac1\pi\langle f,\cos\rangle,$$the map $f\mapsto a_1$ is continuous. A similar argument shows that $f\mapsto b_1$ is continuous too and so your map is the sum of two continuous functions.

Answer 2

By Parseval's Theorem we have that $$\sum_{n \in \Bbb{Z}}|\hat{f}(n)|^2=||f||_2^2$$ or $$\frac{a_0^2}{2}+\sum_{n=1}^{\infty}a_n^2+b_n^2=||f||_2^2$$

So $|T(f)|^2 \leq 2(|a_1|^2+|b_1|^2) \leq \frac{2a_0^2}{2}+2\sum_{n=1}^{\infty}a_n^2+b_n^2=2||f||_2^2$

Thus the operator is bounded.