Semi-Inner Product/Examples/Sequences with Finite Support
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
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.2$