# Inner product is jointly continuous

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 $| \langle x_n, y_n \rangle - \langle x, y \rangle | < \varepsilon$?

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Jointly continuous means, that $\langle, \rangle\colon V^2 \to \mathbb K$ is continuous, right. 2) What are the open balls of your norm? Compare them with products of open balls in $V$. 3) Just use addition of a zero: $\langle x_n, y_n\rangle - \langle x,y \rangle = \langle x_n - x, y\rangle + \langle x_n, y_n - y\rangle$ and Cauchy-Schwarz –  martini Jul 16 '12 at 10:47
Kelley's definition can be applied here but in order to really capture the point here you'd need to do it a bit differently. Be that as it may, I think it's better to keep it down to earth and consider a single function $f : X \times Y \to Z$. Then $f$ is said to be jointly continuous if it's continuous with respect to the product topology on $X \times Y$. The idea is that this is in contrast to the (a priori) weaker condition of separate continuity which means that the functions $f(x,\cdot)\colon Y \to Z$ and $f(\cdot,y)\colon X \to Z$ are continuous for all fixed $x \in X$ and $y \in Y$. –  t.b. Jul 16 '12 at 11:19
@t.b. Yes, thank you, that's what I thought. –  Rudy the Reindeer Jul 16 '12 at 13:29
I don't really think this whole discussion of product topology was necessary, since the problem statement said exactly what is to be proved... The proof does not require any new ideas compared to the proof of the Product rule for limits. –  user31373 Jul 17 '12 at 4:22

$$\lVert ab- a'b' \rVert \le \lVert a \rVert \lVert b-b' \lVert + \lVert a-a'\rVert \lVert b' \rVert$$
That the multiplication operation $(a,b) \mapsto ab$ is jointly continuous in a normed algebra.