Inner Product is Continuous

Theorem
Let $V$ be a inner product space with inner product $\innerprod \cdot \cdot$.

Let $\norm \cdot$ be the inner product norm of $V$.

Define a function $f : V \times V \to \C$ by:


 * $\map f {x, y} = \innerprod x y$

for each $x, y \in V$.

Then $f$ is continuous.

Proof
Let $x, y \in V$.

From Sequential Continuity is Equivalent to Continuity in Normed Vector Space, it suffices to show that:


 * for all sequences $\sequence {x_n}$ and $\sequence {y_n}$ with $x_n \to x$ and $y_n \to y$ we have $\innerprod {x_n} {y_n} \to \innerprod x y$.

That is:


 * for all $\epsilon > 0$, there exists $N \in \N$ such that $\size {\innerprod {x_n} {y_n} - \innerprod x y} < \epsilon$ for $n > N$.

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 for all sequences $\sequence {x_n}$ and $\sequence {y_n}$ with $x_n \to x$ and $y_n \to y$ we have:


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

hence the claim.