Inner Product/Examples/Sequences with Finite Support

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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:


$(1)$ Conjugate Symmetry

\(\ds \innerprod {\sequence {a_n} } {\sequence {b_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map f n a_n \overline { b_n }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \overline {\map f n \overline {a_n} b_n }\) Complex Conjugation is Involution, $\map f n$ is wholly real
\(\ds \) \(=\) \(\ds \overline{ \sum_{n \mathop = 1}^\infty \map f n b_n \overline {a_n} }\) Sum of Complex Conjugates
\(\ds \) \(=\) \(\ds \overline{ \innerprod {\sequence {b_n} } {\sequence {a_n} } }\)

$\Box$


$(2)$ Sesquilinearity

\(\ds \innerprod {\sequence { \lambda a_n + b_n } } {\sequence {c_n} }\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map f n \paren{ \lambda a_n + b_n } \overline {c_n}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren{ \map f n \lambda a_n \overline{c_n} } + \paren{ \map f n b_n \overline {c_n} }\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \map f n \lambda a_n \overline{c_n} + \sum_{n \mathop = 1}^\infty \map f n b_n \overline {c_n}\)
\(\ds \) \(=\) \(\ds \lambda \sum_{n \mathop = 1}^\infty \map f n a_n \overline{c_n} + \sum_{n \mathop = 1}^\infty \map f n b_n \overline {c_n}\)
\(\ds \) \(=\) \(\ds \lambda \innerprod {\sequence {a_n} } {\sequence {c_n} } + \innerprod {\sequence {b_n} } {\sequence {c_n} }\)

$\Box$


$(3)$ Non-Negative Definiteness

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

$\Box$


$(4)$ 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 $n \in \N$.

$\Box$


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

$\blacksquare$


Sources

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