Semi-Inner Product/Examples/Sequences with Finite Support

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Example of Semi-Inner Product

Let $\GF$ be a subfield of $\C$.

Let $V$ be the vector space of sequences with finite support over $\GF$.


Let $\innerprod \cdot \cdot: V \times V \to \GF$ be the mapping defined by:

$\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty a_{2 n} \overline {b_{2 n} }$


Then $\innerprod \cdot \cdot$ is a semi-inner product on $V$ but not an inner product on $V$.


Proof

First of all, note that $V$ contains only the sequences with finite support.

Therefore, for each $\sequence {a_n}, \sequence{b_n}$ there exists $N \in \N$ such that:

$\forall n \ge N: a_n = b_n = 0$

and hence:

$\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty a_{2 n} \overline {b_{2 n} } = \sum_{n \mathop = 1}^{N / 2} a_{2 n} \overline {b_{2 n} }$

so that $\innerprod \cdot \cdot: V \times V \to \GF$ is indeed defined.


Now checking the axioms for a semi-inner product in turn:


$(1)$ Conjugate Symmetry

\(\ds \innerprod {\sequence {a_n} } {\sequence {b_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty a_{2 n} \overline { b_{2 n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \overline {\overline {a_{2 n} } b_{2 n} }\) Complex Conjugation is Involution
\(\ds \) \(=\) \(\ds \overline{ \sum_{n \mathop = 1}^\infty b_{2 n} \overline {a_{2 n} } }\) Sum of Complex Conjugates
\(\ds \) \(=\) \(\ds \overline{ \innerprod {\sequence {b_n} } {\sequence {a_n} } }\)

$\Box$


$(2)$ Sesquilinearity

\(\ds \innerprod {\sequence { \lambda a_n + b_n } } {\sequence {c_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren{ \lambda a_{2 n} + b_{2_n} } \overline {c_{2 n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren{ \lambda a_{2 n} \overline{c_{2 n} } } + \paren{ b_{2 n} \overline {c_{2 n} } }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \lambda a_{2 n} \overline{c_{2 n} } + \sum_{n \mathop = 1}^\infty b_{2 n} \overline {c_{2 n} }\)
\(\ds \) \(=\) \(\ds \lambda \sum_{n \mathop = 1}^\infty a_{2 n} \overline{c_{2 n} } + \sum_{n \mathop = 1}^\infty b_{2 n} \overline {c_{2 n} }\)
\(\ds \) \(=\) \(\ds \lambda \innerprod {\sequence {a_n} } {\sequence {c_n} } + \innerprod {\sequence {b_n} } {\sequence {c_n} }\)

$\Box$


$(3)$ Non-Negative Definiteness

\(\ds \innerprod {\sequence {a_n} } {\sequence {a_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty a_{2 n} \overline {a_{2 n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \cmod{ a_{2 n} }^2\) Product of Complex Number with Conjugate
\(\ds \) \(\in\) \(\ds \R_{\ge 0}\)

$\Box$

Hence $\innerprod \cdot \cdot$ is a semi-inner product.


Because any sequence $\sequence{a_n}$ such that $a_{2n} = 0$ for all $n \in \N$ will have:

$\innerprod {\sequence{a_n} } {\sequence{a_n} } = 0$

it follows that $\innerprod \cdot \cdot$ is not an inner product.

$\blacksquare$


Sources

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