Inner Product/Examples/Sequences with Finite Support

Example of Inner Product
Let $\GF$ be either $\R$ or $\C$.

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

Let $f: \N \to \R_{>0}$ be a mapping.

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


 * $\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty \map f n a_n \overline{ b_n }$

Then $\innerprod \cdot \cdot$ is 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:


 * $\ds \innerprod {\sequence {a_n} } {\sequence {b_n} } = \sum_{n \mathop = 1}^\infty \map f n a_n \overline {b_n} = \sum_{n \mathop = 1}^{N-1} \map f n a_n \overline {b_n }$

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

Now checking the axioms for an inner product in turn:

$(4)$ Positivity
Suppose that $\innerprod { \sequence{a_n} } { \sequence{a_n} } = 0$.

That is:


 * $\ds \sum_{n \mathop = 1}^\infty \map f n \cmod{ a_n }^2 = 0$

For each $n \in \N$, we have $\map f n \cmod{ a_n }^2 \ge 0$.

Hence, for each $n \in \N$:


 * $\ds \sum_{n \mathop = 1}^\infty \map f n \cmod{ a_n }^2 \ge \map f n \cmod{ a_n }^2$

Thus for each $n \in \N$:


 * $\map f n \cmod{ a_n }^2 = 0$

Since $\map f n > 0$ it follows that $\cmod{ a_n }^2 = 0$.

Therefore $a_n = 0$ for all $n \in \N$.

Having verified all the axioms, we conclude $\innerprod \cdot \cdot$ is an inner product.