Consider the following linear operators $G_k:\ell_2\to \ell_2$
a) $G_1 : x \mapsto (x_1+x_2+x_3,x_4,x_5,\dots)$
b) $G_2 : x \mapsto (x_1-4x_2+3x_3,x_4,x_5,\dots)$
c) For fixed $(z_1,z_2,z_3)\in \mathbb{R}^3$, $G_3 : x \mapsto (z_1x_1,z_2x_2,z_3x_3,x_4,x_5,\dots)$
I have been asked to prove that the above linear operators $G_1, G_2, G_3$ are bounded linear operators.
Attempts
a) Let $x\in \ell_2$, $x = (x_1,x_2,x_3,\dots)$
\begin{align*}
\Vert G_1\Vert_2^2 &= \Vert (x_1+x_2+x_3,x_4,x_5,\dots)\Vert_2^{2} \\
& = (x_1+x_2+x_3)^2+x_4^2+x_5^2 \\
&\leq x_1^2+x_2^2+x_3^2+x_4^2+\cdots \\
& = \Vert {x_n}\Vert_2^2
\end{align*}
b) $x\in \ell_2$, $x = (x_1,x_2,x_3,\dots)$
\begin{align*}
\Vert G_2\Vert_2^2 &= \Vert (x_1-4x_2+3x_3,x_4,x_5,\dots)\Vert_2^{2} \\
& = (x_1-4x_2+3x_3)^2+x_4^2+x_5^2 \\
& \leq x_1^2+16x_2^2+9x_3^2+x_4^2+\cdots
\end{align*}
c) $x\in \ell_2$, $x = (x_1,x_2,x_3,\dots)$
\begin{align*}
\Vert G_3\Vert_2^2 &= \Vert (z_1x_1,z_2x_2,z_3x_3,x_4,x_5,\dots)\Vert_2^{2} \\
&= z_1^2x_1^2+z_2^2x_2^2+z_3^2x_3^2+x_4^2+\cdots
\end{align*}
I am not sure about a). For b) and c) I am stuck in showing that the last steps are bounded by $C\Vert x \Vert_2^2$.
Best Answer
The idea is to show that for each of the operators $G_j$ defined in the OP, there are contants $k_j$ such that $$\|Gj\mathbf{x}\|_2\leq k_j\|\mathbf{x}\|_2$$ where $\mathbf{x}=[x_1,x_2,\ldots,]\in \ell_2$.
(a): Let $\mathbf{x}=(x_1,x_2,x_3,x_4,\ldots)$. Then $$\begin{align} G\mathbf{x}&=(x_1+x_2+x_3,x_4,x_5,\ldots)\\ &=(x_1+x_2+x_3,0,0,\ldots)+(0,x_4,x_5,\ldots)\\ &=\mathbf{a}+\qquad\qquad\qquad\qquad+\mathbf{b} \end{align}$$ Hence $$\begin{align} \|G\mathbf{x}\|_2&\leq \|\mathbf{a}\|_2+\|\mathbf{b}\|_2\leq|x_1|+|x_2|+|x_3|+\|\mathbf{b}\|_2\\ &\leq \sqrt{3}\sqrt{|x_1|^2+|x_2|^2+|x_3|^2}+\|\mathbf{b}\|_2\\ &\leq(\sqrt{3}+1)\|\mathbf{x}\|_2 \end{align} $$ Since $\sqrt{|x_1|^2+|x_2|^2+|x_3|^2}\leq\|\mathbf{x}\|_2$ and $\|\mathbf{b}\|_2\leq \|\mathbf{x}\|_2$.
(b) and (c) can be dealt with in a similar manner. For example, in (c) one can easily obtain that $$\|G_3\mathbf{x}\|_2\leq \max(1,|z_1|,|z_2|,|z_3|)\|\mathbf{x}\|_2$$