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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    $\begingroup$ 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 $\endgroup$
    – martini
    Jul 16, 2012 at 10:47
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    $\begingroup$ 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 $y \in Y$. $\endgroup$
    – t.b.
    Jul 16, 2012 at 11:19
  • $\begingroup$ @t.b. Yes, thank you, that's what I thought. $\endgroup$ Jul 16, 2012 at 13:29
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    $\begingroup$ 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. $\endgroup$
    – user31373
    Jul 17, 2012 at 4:22

2 Answers 2


We will use the infinity or max norm on $V\times V$, that is, for $(u,v)\in V\times V$, $\lVert (u,v)\rVert_{V\times V}=\max\{\lVert u\rVert_V,\lVert v\rVert_V\}$, where the norm on $V$ is the norm induced by the inner product $\langle\cdot,\cdot\rangle$. I can't speak with authority on why this is equivalent to the product topology, but I don't think it should be too difficult to show for a finite product of normed spaces - those are some well-behaved constructs. For a function to be jointly continuous, it means simply that it is continuous in the product topology, rather than only being continuous in each variable separately, with the others fixed. That is, for $f:X\times Y\to Z$, $f$ is jointly continuous if, for all $(x_0,y_0)\in X\times Y$, given any $\epsilon>0$, there exists a $\delta>0$ such that, for all $(x,y)\in X\times Y$, $d_{X\times Y}((x,y),(x_0,y_0))<\delta$ implies $d_Z(f(x,y),f(x_0,y_0))<\epsilon$. So let's do it, using the induced norm as our metric.

Fix an $x_0,y_0\in V$, and let $\epsilon>0$ be given. Choose $\delta=\min\{1,\frac{\epsilon}{2(\lVert y_0\rVert + 1)},\frac{\epsilon}{2(\lVert x_0\rVert+1)}\}$, and note $\lvert\lVert y\rVert_V-\lVert y_0\rVert_V\rvert\leq \lVert y-y_0\rVert_V<1$, by the reverse triangle inequality and our choice of $\delta$. This implies $\lVert y\rVert_V<1+\lVert y_0\rVert_V$.

Let $x,y\in V$, and suppose $\lVert (x,y)-(x_0,y_0)\rVert_{V\times V}=\lVert(x-x_0,y-y_0)\rVert_{V\times V}<\delta$, implying $\lVert x-x_0\rVert_V<\delta$ and $\lVert y-y_0\rVert_V<\delta$. Then

$\lvert\langle x,y\rangle-\langle x_0,y_0\rangle\rvert = \lvert\langle x,y\rangle-\langle x_0,y\rangle+\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$

$\leq \lvert\langle x,y\rangle-\langle x_0,y\rangle\rvert+\lvert\langle x_0,y\rangle-\langle x_0,y_0\rangle\rvert$

$= \lvert\langle x-x_0,y\rangle\rvert+\lvert\langle x_0,y-y_0\rangle\rvert$

$\leq \lVert x-x_0\rVert_V\lVert y\rVert_V + \lVert x_0\rVert_V \lVert y-y_0\rVert_V$ (by Cauchy-Bunyakovsky-Schwarz inequality)

$< (\frac{\epsilon}{2(\lVert y_0\rVert+1)})(1+\lVert y_0\rVert_V) + (\lVert x_0\rVert_V+1)(\frac{\epsilon}{2(\lVert x_0\rVert+1)})$, (since $\lVert x_0\rVert_V<\lVert x_0\rVert_V+1$, and by our choice of $\delta$)

$= \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$.


I'm not quite sure if this helps, but reading G Murphy's C* Algebras and Operator Theory he says that it is evident from the following inequality

$$\lVert ab- a'b' \rVert \le \lVert a \rVert \lVert b-b' \lVert + \lVert a-a'\rVert \lVert b' \rVert$$

That the multiplication operation $(a,b) \mapsto ab$ is jointly continuous in a normed algebra.


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