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Left shift operator is $L:\ell^2\to\ell^2$ defined by $$(x_1,x_2,x_3,x_4,\ldots)\mapsto (x_2,x_3,x_4,\ldots).$$

This is not an isometry apparently, so $\|Lx\|\ne \|x\|$.

Does this mean $\|L\|\ne1$?

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  • $\begingroup$ L is an operator between wich spaces? $L: L_\infty \to L_\infty$? $\endgroup$ Jun 23, 2015 at 17:13

2 Answers 2

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No, Consider $L: \ell_2 \to \ell_2$. Since $\vert \vert L \vert \vert_{op} = \sup_{\vert \vert x \vert \vert_2 =1} \vert \vert L(x) \vert \vert_2 $

Since $\vert \vert L(x) \vert \vert_2^2 = \sum_{i \geq 2} x_i^2$ and $\vert\vert x \vert\vert_2 = \sum_{i \geq 1} x_i^2$ it is clear that $ \vert \vert L(x) \vert \vert_2 \leq \vert \vert x \vert \vert_2^2 $

but take $e_2=(0,1,0,0,\ldots)$, $L(e_2) = (1,0,0, \ldots)$ $$\vert \vert L(e_2) \vert \vert_2 = 1 = \vert \vert e_2 \vert \vert_2 $$

so $\vert\vert L \vert\vert_{op} = 1$

remark: $\ell_2 = \{x = (x_1, x_2, \ldots) \mid x_i \in \mathbb{R}, \sum_{i \geq 1}x_i^2 < \infty\}$

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  • $\begingroup$ Beat me to it :) Note that you can write \| instead of \vert\vert $\endgroup$
    – Math1000
    Jun 23, 2015 at 17:26
  • $\begingroup$ Actually, it's l^2. Sorry. $\endgroup$
    – user250153
    Jun 23, 2015 at 17:34
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It is the adjoint of right shift operator which is a isometry, so it's norm is 1.

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  • $\begingroup$ Perhaps if you can elaborate on why the right shift operator is an isometry, that might help the OP out a bit more. $\endgroup$
    – Ken
    Jun 23, 2015 at 18:14
  • $\begingroup$ books.google.co.in/…^2+to+l^2+is+an+linear+isometry&source=bl&ots=qVcsciAMTQ&sig=P53yQIxyPnusG_-y2DE1ZRwvMSc&hl=en&sa=X&ei=1SGKVYP5AdCbuQTJ8oK4BQ&ved=0CCkQ6AEwAw#v=onepage&q=right%20shift%20operator%20on%20l^2%20to%20l^2%20is%20an%20linear%20isometry&f=false $\endgroup$ Jun 24, 2015 at 3:20
  • $\begingroup$ first check this link then tell. $\endgroup$ Jun 24, 2015 at 3:21

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