Bounded linear operator and self adjoint operator these questions are in my workbook but there is no worked solutions whatsoever. I dont know where to begin with this at all. Any help would be much appreciated. Thankyou

 A: $M_c(x) = (c_1 x_1, c_2 x_2,...)$. It is easy to see that
$\| M_c (x) \|^2 = \sum_k |c_k|^2 |x_k|^2  \le \sum_k |x_k|^2 = \|x\|_2$,
and so $\|M_c\| \le 1$. Noting that $\|M_c (e_n)\| = |c_n|$, shows that $\|M_c \| = 1$.
If $y = M_c x$, then we have $y_k = c_k x_k$, so if it is invertible, we must have
$x_k = {1 \over c_k} y_k$, that is, the inverse will be the multiplication operator with value $\kappa = ({1 \over c_1}, {1 \over c_2},...)$. To be invertible, we need $c_k \neq 0$ for all $k$, otherwise $M_c e_n = 0$ for some $n$, which would mean that $M_c$ is not invertible.
Since $|c_k|$ is increasing and bounded above, this implies $c_1 \neq 0 $.
It is easy to see that if $c_1 \neq 0$, then ${1 \over |c_k|} \le {1 \over |c_1|}$ for all $k$, hence the operator $A(x) = ({1 \over c_1}x_1, {1 \over c_2} x_2,...)$ is continuous and satisfies $A( M_c(x)) = x$ for all $x$, hence $M_c$ is invertible.
To be self-adjoint, we need $M_c = M_c^*$.
This is true iff $\langle e_i, M_c e_j \rangle = \langle M_c e_i,  e_j \rangle $
for all $i,j$ iff ${ 1 \over c_j } \delta_{ij} = { 1 \over \bar{c_i} } \delta_{ij} $ for all $i,j$ iff $\bar{c_i} = c_i$ for all $i$ iff $c_i \in \mathbb{R}$ for all $i$.
A: For the first problem, note that
$$
\|M_c\| = \sup_{\|x\|_2=1} \left(\sum_{i=1}^\infty |c_n x_n| \right)^{1/2}
$$
It is fairly easy to show that $1$ gives us an upper bound for this infinite sum.  Why is it also the least-upper bound?
For the second: note that if $M_c$ is invertible, its inverse would have to be $M_d$ with $d = \{\frac 1{c_n}\}_{n=1}^\infty$
For the third: as a means of intuition, when is a diagonal matrix self-adjoint?
