Inner Product is Continuous

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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.

$\blacksquare$