Norm and boundedness of a linear operator

Question

Let $\ell^2 = {(x_1,x_2,..), x_n \in \mathbb{C} ; \sum^\infty_{i=1}|x_i|^2<\infty}$ and let $\|(x_1,x_2,..)\| = \sqrt{\sum_{i=1}^\infty|x_i|^2} $. Operator $M : \ell^2 \rightarrow \ell^2$ is defined as $M(x_1,x_2,x_3,\ldots)= (x_2,x_2,x_3,\ldots)$. Can someone help please and explain how to show that $M$ is bounded and how to find its norm? Thanks in advance

So we need to find C, sucht that $||Ax||\leq C||x||$. So $||Ax|| = ||(x_2,x_2,x_3..)|| = \sqrt{\sum_{i=2}^\infty|x_i| + |x_2|^2}$ and how to proceed next?

Answer 1

$\|M(x)\|=\sqrt{|x_2|^2+\|x\|^2-|x_1|^2}\leq \sqrt{2}\|x\|$ implies that $M$ is bounded, let $e_2=(0,1,0...0,..)$, $M(e_2)=(1,1,0,..0,..)$ implies that $\|M(e_1)\|=\sqrt2$, thus the norm is $\sqrt2$.

Answer 2

Hint Let $$T(x_1,x_2,x_3,x_4,..)= (x_2,0,0,..,0, ..) \\ S(x_1,x_2,x_3,...)=(0,x_2,x_3,....)$$

Show that both $T,S$ are bounded and use the fact that $M=T+S$ to deduce that $M$ is bounded.