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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.

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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.

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    $\begingroup$ Note that $a_1(f_1+f_2)=\frac1\pi\langle f_1+f_2,\cos\rangle=\frac1\pi\langle f_1,\cos\rangle+\frac1\pi\langle f_2,\cos\rangle=a_1(f_1)+a_1(f_2)$ and that $a_1(\lambda f)=\frac1\pi\langle\lambda f,\cos\rangle=\lambda\frac1\pi\langle f,\cos\rangle=\lambda a_1(f)$. $\endgroup$ – José Carlos Santos Dec 11 '19 at 15:45
  • $\begingroup$ Dear Jose, really thanks again! I deleted comment because I understand why it is linear and I am sorry for bothering you with such silly questions. $\endgroup$ – Alpys Rauan Dec 11 '19 at 17:29
  • $\begingroup$ If you do not mind, can you look at continuation of this problem? I posted separately, math.stackexchange.com/questions/3472635/… $\endgroup$ – Alpys Rauan Dec 11 '19 at 17:30
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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.

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