Operator between Hilbert spaces, boundness, image and eigenvalues I'm totally new in functional analysis and this is my first problem. Let's $H=L^2(-\pi,\pi)$ as Hilbert space with basis $u_n+iv_n$ where
$$u_0 = \frac{1}{\sqrt{2\pi}},\,\,u_n(x)=\frac{1}{\sqrt{\pi}}\cos(nx)$$
$$v_n(x) =\frac{1}{\sqrt{\pi}}\sin(nx)$$
and define the following operator 
$$Tu_0=0,\,\,Tu_n =\frac{v_n}{n},\,\,Tv_n=\frac{u_n}{n}$$
It's this operator bounded? And it's the image dense on $H$? What about his eigenvalues?
Thank you very much for the help, I'm not very fond of my analysis skills.
 A: The set $\{ u_0, v_1, u_1, v_2, u_2, \cdots \}$ is a complete orthonormal basis of $L^2(-\pi,\pi)$. So there is an isometric correspondence between $L^2(-\pi,\pi)$ and $\ell^2(\mathbb{N})$ defined by
$$
          f \sim ((f,u_0),(f,v_1),(f,u_1),(f,v_2),(f,u_2),\cdots).
$$
This is due to Parseval's equality.
Using this correspondence, the operator $T$ looks like the following on $\ell^2(\mathbb{N})$:
$$
         \tilde{T}(a_0,b_1,a_1,b_2,a_2,b_3,a_3,\cdots)=(0,a_1,b_1,\frac{1}{2}a_2,\frac{1}{2}b_2,\frac{1}{3}a_3,\frac{1}{3}b_3,\cdots).
$$
You can see this operator is bounded because
\begin{align}
       \|\tilde{T}(a_0,b_1,a_1,b_2,a_2,\cdots)\|^2 &=a_1^2+b_1^2+\frac{1}{4}a_2^2+\frac{1}{4}b_2^2+\frac{1}{9}a_3^2+\frac{1}{9}b_3^2+\cdots \\
       & \le a_0^{2}+a_1^2+b_1^2+a_2^2+b_2^2+\cdots \\
       & = \|(a_0,b_1,a_1,b_2,a_2,\cdots)\|_{\ell^2}^2
\end{align}
In other words, $\|Tf\|_{L^2}\le \|f\|_{L^2}$. The range of $T$ does not include $u_0$. In fact, $\mathcal{R}(T)\perp u_0$, which guarantees that the range of $T$ is not dense. It's easy to verify that
\begin{align}
        Tu_0 & = 0 \\
        T(u_n+v_n)&=\frac{1}{n}(u_n+v_n), \;\;\; n=1,2,3,\cdots, \\
        T(u_n-v_n)&=-\frac{1}{n}(u_n-v_n),\;\;\; n=1,2,3,\cdots.
\end{align}
So $0,\pm 1,\pm\frac{1}{2},\pm\frac{1}{3},\cdots$ are eigenvalues.
