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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 $C\Vert x \Vert_2^2$.

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    $\begingroup$ You need to show they are bounded by $C\Vert x\Vert^2$ for some positive $C$. $\endgroup$ Commented Mar 18, 2022 at 17:09
  • $\begingroup$ For example in b) and c), I am struggling to deduce that fact $\endgroup$
    – user758734
    Commented Mar 18, 2022 at 17:10
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    $\begingroup$ What you wrote for $a$ isn't correct. $\endgroup$ Commented Mar 18, 2022 at 17:12
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    $\begingroup$ 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 $b$. $\endgroup$ Commented Mar 18, 2022 at 17:15
  • $\begingroup$ I am trying to find C, I have edited the question if you have not noticed. $\endgroup$
    – user758734
    Commented Mar 18, 2022 at 17:26

2 Answers 2

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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 $$\|G_3\mathbf{x}\|_2\leq \max(1,|z_1|,|z_2|,|z_3|)\|\mathbf{x}\|_2$$

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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 $G_3$).

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