Norm of a left shift operator 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 $\|L\|\ne1$?
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: $\ell_2 = \{x = (x_1, x_2, \ldots) \mid x_i \in \mathbb{R}, \sum_{i \geq 1}x_i^2 &lt; \infty\}$
It is the adjoint of right shift operator which is a isometry, so it's norm is 1.