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:

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:

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:

So we have:


 * $\innerprod {x_n} {y_n} \to \innerprod x y$

as $n \to \infty$, hence the claim.