Right Shift Operator on 2-Sequence Space is Continuous

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Theorem

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence space with $2$-norm.

Let $R : \ell^2 \to \ell^2$ be the right shift operator.


Then $R$ is continuous on $\struct {\ell^2, \norm {\, \cdot \,}_2}$.


Proof

Let $\sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, a_3, \ldots}$ be a $2$-sequence.

\(\ds \norm {\map R {\sequence {a_n}_{n \mathop \in \N} } }_2\) \(=\) \(\ds \norm { R \tuple {a_1, a_2, a_3, \ldots} }_2\)
\(\ds \) \(=\) \(\ds \norm {\tuple {0, a_1, a_2, \ldots} }_2\) Definition of Right Shift Operator
\(\ds \) \(=\) \(\ds \sqrt {0^2 + \sum_{i \mathop = 1}^\infty \size {a_i}^2}\)
\(\ds \) \(=\) \(\ds \sqrt {\sum_{i \mathop = 1}^\infty \size {a_i}^2}\)
\(\ds \) \(=\) \(\ds 1 \cdot \norm {\sequence {a_n}_{n \mathop \in \N} }_2\)

By continuity of linear transformations, $R$ is continuous in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.

$\blacksquare$


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