I'm attempting another exercise from my notes:
Show that an inner product on an inner product space is jointly continuous with respect to the induced norm:if $v_n \to v$ and $w_n \to w$ as $n \to \infty$, then $\langle v_n, w_n\rangle \to \langle v,w \rangle$ as $n \to \infty$.
I'd not heard jointly continuous before so I googled and found the following definition in Kelley:
Let $P: F \times X \to Y$ defined by $(f,x) \mapsto f(x)$. Each topology on $F$ gives rise to a product topology on $F \times X$. A topology for $F$ is said to be jointly continuous iff $P$ is continuous.
I think in the case of the inner product, $F$ is the one point space $\{ \langle \cdot, \cdot \rangle \}$ and $X = V \times V$. Then there is only one topology on $F$ and jointly continuous just means that the inner product is continuous. Question 1: Is that correct so far?
So to show that $\langle \cdot, \cdot \rangle$ is continuous we need to show that if $(x_n, y_n) \to (x,y) $ then $\langle x_n, y_n \rangle \to \langle x, y \rangle$ in $\mathbb R$.
On $V \times V$ we can define the norm $\|(a,b)\| = \max(\|a\|, \|b\|)$. Question 2: How do I show that this norm induces the same topology as the product topology?
Now I want to show that for $\varepsilon > 0$ there is an $N$ such that for $n > N$, $| \langle x_n, y_n \rangle - \langle x, y \rangle | < \varepsilon$. In the max norm, I have $\|x_n - x\|^2 = \langle x_n, x_n \rangle - \langle x_n, x \rangle - \langle x, x_n \rangle + \langle x, x \rangle < \varepsilon^2$, for both $x_n,x$ and $y_n,y$. Question 3: How can I use this to show $| \langle x_n, y_n \rangle - \langle x, y \rangle | < \varepsilon$?