Inner Product is Continuous
Theorem
Let $\struct {V, \innerprod \cdot \cdot}$ be a inner product space.
Let $\norm \cdot$ be the inner product norm on $V$.
Let $x, y \in V$.
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging in $\struct {V, \norm \cdot}$ to $x$ and $y$ respectively.
Then we have:
- $\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$.
Proof
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging to $x$ and $y$ respectively.
From Modulus of Limit in Normed Vector Space, we have that:
- $\norm {x_n} \to \norm x$
and:
- $\norm {y_n} \to \norm y$
We have:
\(\ds \size {\innerprod {x_n} {y_n} - \innerprod x y}\) | \(=\) | \(\ds \size {\innerprod {x_n - x} {y_n} + \innerprod x {y_n - y} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {\innerprod {x_n - x} {y_n} } + \size {\innerprod x {y_n - y} }\) | Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n - x} \norm {y_n} + \norm x \norm {y_n - y}\) | Cauchy-Bunyakovsky-Schwarz Inequality |
From Convergent Real Sequence is Bounded, we have:
- $\sequence {\norm {y_n} }$ is bounded.
That is, there exists a positive real number $M$ such that:
- $\norm {y_n} < M$
Let $\epsilon > 0$.
From the definition of a convergent sequence in a normed vector space, there exists $N_1 \in \N$ such that:
- $\norm {x_n - x} < \dfrac \epsilon {2 M}$
for $n > N_1$.
If $x = 0$, we have:
- $\norm x = 0$
Then:
\(\ds \size {\innerprod {x_n} {y_n} - \innerprod x y}\) | \(\le\) | \(\ds \norm {x_n - x} \norm {y_n}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds M \norm {x_n - x}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
for $n > N_1$, so we are done in this case.
Now take $x \ne 0$, then:
- $\norm x \ne 0$
So by the definition of a convergent sequence in a normed vector space, there exists $N_2 \in \N$ such that:
- $\norm {y_n - y} < \dfrac \epsilon {2 \norm x}$
Let:
- $N = \max \set {N_1, N_2}$
Then, for $n > N$, we have:
\(\ds \size {\innerprod {x_n} {y_n} - \innerprod x y}\) | \(\le\) | \(\ds \norm {x_n - x} \norm {y_n} + \norm x \norm {y_n - y}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds M \times \frac \epsilon {2 M} + \norm x \times \frac \epsilon {2 \norm x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
So we have:
- $\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$, hence the claim.
$\blacksquare$