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