Characterizing image of operator between Hilbert spaces Let be $\{u_0,v_1,u_1,\ldots\}$ the Fourier basis of $L^2(0,\pi)$ with $u_0=(2\pi)^{-1/2}$, $u_n(x)=(\pi)^{-1/2}\cos(nx)$, $v_n(x)=(\pi)^{-1/2}\sin(nx)$ and let be $A$ the operator defined by $$Au_0=0,\space\space Au_n=v_n/n,\space\space Av_n=u_n/n$$
which acts on $\ell^2({\mathbb{N}})$ as
$$A(a_0,b_1,a_1,b_2,\ldots)=(0,a_1,b_1,a_2/2,b_2/2,\ldots$$
This is a bounded operator and we can find eigenvectors:
$$A(u_n\pm v_n)=\pm(u_n\pm v_n)/n$$
Now, the questions:


*

*Can we conclude that the operator is self-adjoint? How are we sure those are the only eigenvector?

*Is the operator compact?
And the most important one:


*

*Are the functions in $A(L^2(0,\pi))$ continuous?


The last statement is the most intriguing: if I write every $f\in L^2(0,\pi)$  in Fourier series
$$f=\sum_{n\geq 0} (\alpha_n u_n+\beta_n v_n)$$ 
then 
$$Af=\sum_{n\geq 0}\frac {\alpha_n v_n + \beta_n u_n}{n}$$
If I prove uniform convergence of this series I'm done, but I'm not sure how to bound Fourier coefficients.
 A: The set of eigenvectors $\mathcal{E}=\{ e_0,e_1^-,e_1^+,e_2^-,e_2^+,\cdots\}$ is a complete orthonormal basis of $L^2$, where
$$
        e_0, e_1^-=\frac{1}{\sqrt{2}}(u_1-v_1),e_1^+=\frac{1}{\sqrt{2}}(u_1+v_1),\cdots.
$$
$\mathcal{E}$ is complete because every $u_j,v_j$ can be written in terms of the elements of $\mathcal{E}$. The operator $A$ is diagonal in this basis. $A$ is a bounded linear operator. In this basis, it is easy to check that $(Af,g)=(f,Ag)$ for all $f,g\in L^2$, which proves that $A$ is selfadjoint.
$A$ cannot have other eigenvectors because, if $Af=\lambda f$, then $f$ has the representation $(f_0,f_1^-,f_1^+,f_2^-,f_2^+,\cdots)$ in this basis $\mathcal{E}$, which gives
$$
          (0,f_1^-,f_1^+,\frac{1}{2}f_2^-,\frac{1}{2}f_2^+,\cdots)=(\lambda f_0,\lambda f_1^-,\lambda f_1^+,\lambda f_2^-,\lambda f_2^+,\cdots),\\
          (0-\lambda f_0,(1-\lambda)f_1^-,(1-\lambda)f_1^+,(1/2-\lambda)f_2^-,(1/2-\lambda)f_2^+,\cdots)=0.
$$
If $\lambda\notin\{0,1,1/2,1/3,\cdots\}$, then all coefficients for $f$ are $0$. So $\{ 0,1,1/2,1/3,\cdots\}$ is the set of eigenvaues. If $\lambda =0$, then $f=f_0(1,0,0,0,\cdots)$. If $\lambda=1/n$ for some $n\in \{1,2,\cdots\}$, then the above forces $f=f_n^-e_n^-+f_n^+e_n^+$.
The functions in the range of $A$ are continuous because
$$
       Af=\sum_{n=1}^{\infty}\frac{1}{n}\{(f,e_n^-)e_n^-+(f,e_n^+)e_n^+\}
$$
not only converges in $L^2$, but it is absolutely convergent by the Weierstrass $M$ test because of the pointwise uniform estimates
$$
       \left|\frac{1}{n}\{(f,e_n^-)e_n^-+(f,e_n^+)e_n^+\}(x)\right|
        \le \frac{\sqrt{2}}{n}\{|(f,e_n^-1)|+|(f,e_n^+)|\}.
$$
Indeed, the right side does not depend on $x$ and the sum of all the terms on the right is bounded, which is seen by an application of the Cauchy-Schwarz inequality and the fact that
$$
               \sum_{n=0}^{\infty}|(f,e_n^-)|^2+|(f,e_n^+)|^2 = \|f\|^2.
$$
