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:

$(3)$ Non-Negative Definiteness
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.