# Bounded Linear Operators in $\ell_2$

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 $$C\Vert x \Vert_2^2$$.

• You need to show they are bounded by $C\Vert x\Vert^2$ for some positive $C$. Commented Mar 18, 2022 at 17:09
• For example in b) and c), I am struggling to deduce that fact
– user758734
Commented Mar 18, 2022 at 17:10
• What you wrote for $a$ isn't correct. Commented Mar 18, 2022 at 17:12
• You said you're trying to show that the last steps are bounded by $||x||_2^2$, but this is not true. They are bounded by $C||x||_2^2$ for some positive $C$ though. You should be thinking Cauchy-Schwarz for $a$ and $b$. Commented Mar 18, 2022 at 17:15
• I am trying to find C, I have edited the question if you have not noticed.
– user758734
Commented Mar 18, 2022 at 17:26

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 $$\|G_3\mathbf{x}\|_2\leq \max(1,|z_1|,|z_2|,|z_3|)\|\mathbf{x}\|_2$$
It is possible to calculate the norm of these operators. For example $$\|G_1x\|^2= |x_1+x_2+x_3|^2+\sum_{n=4}^\infty |x_n|^2\\ \le 3(|x_1|^2+|x_2|^2+|x_3|^2)+\sum_{n=4}^\infty |x_n|^2\le 3\|x\|^2.$$ Hence $$\|G_1\|\le \sqrt{3}.$$ The norm is attained at $$x=(1,1,1,0,0,\ldots ),$$ i.e. $$\|G_1\|= \sqrt{3}.$$ Similarly $$\|G_2\|=\sqrt{26},$$ and the norm is attained at $$x=(1,-4,3,0,0,\ldots ).$$ The norm $$\|G_3\|$$ is calculated in answer 1 and is attained at one of the basic vectors $$\delta_j$$ for $$j=1,2,3,4.$$
In general the boundedness of all three operators follows from the fact that each of them is of the form $$U+F,$$ where $$U$$ is an isometry on a subspace (of codimension $$3$$) and a one (for $$G_1,\ G_2$$) or three dimensional bounded operator (for $$G_3$$).