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Let $f: \mathbb R \times \mathbb R \to \mathbb R$ be the map $(x,y) \mapsto xy$. For my own exercise I wanted to prove it is continuous. It seemed easy to assume that the topology on $ \mathbb R \times \mathbb R$ is generated by the $\max$-norm. Then let $(x_0,y_0)$ be a fixed point. The goal is to show that $|xy - x_0y_0|$ can be made less than epsilon for points $(x,y)$ appropriately. I did this: $$ |xy-x_0y_0| = |xy-x_0y +x_0y -x_0y_0| \le |y||x-x_0| + |x_0||y - y_0| $$

The problem is, $|y|$ is not constant therefore it is not possible to let $\delta = {\varepsilon \over |y|+|x_0|}$. My question is: how to do it? I have tried but can't do it.

blue
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1 Answers1

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One method is to write

$$xy - x_0y_0 = (x-x_0)y_0 + x_0(y-y_0) + (x-x_0)(y-y_0).$$

Then you can set $$\delta = \frac{\min \{ 1,\, \varepsilon\}}{3(\lvert x_0\rvert + \lvert y_0\rvert + 1)}$$

to have $\lVert (x,y) - (x_0,y_0)\rVert_\infty < \delta \Rightarrow \lvert xy-x_0y_0\rvert < \varepsilon$.

Daniel Fischer
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  • I am sorry: why $\varepsilon \le 1$ is needed? – blue Dec 19 '13 at 16:20
  • Because the last term can be (almost) as large as $\delta^2$, and we want that smaller than $\varepsilon/3$ like the other terms. We could use a somewhat larger bound for $\varepsilon$, $\varepsilon \leqslant 3$ would do here, but $\varepsilon \leqslant 1$ is a habit. It doesn't matter, since we're only interested in small $\varepsilon > 0$, however. Nevertheless, let me change it a bit to get rid of the assumption. – Daniel Fischer Dec 19 '13 at 16:26
  • Thank you, now I understand better. I did the computation and it also seems to work with $$ \delta = \frac{\min { 1,, \varepsilon}}{\lvert x_0\rvert + \lvert y_0\rvert + 1}$$ Btw, it is an amazing trick. I tried for very long and did not manage to find this trick. – blue Dec 19 '13 at 16:45
  • Indeed, you're right. When not considering each term in isolation, we don't need each smaller than $\varepsilon/3$ to know that the sum is smaller than $\varepsilon$, it suffices that we get a bound $\delta(\lvert x_0\rvert + \lvert y_0\rvert + 1)$. Good observation! – Daniel Fischer Dec 19 '13 at 16:50
  • In related link to my question I found this. Here the answer is what I tried to do but it doesn't finish. Is it wrong? – blue Dec 20 '13 at 08:18
  • No, it's not wrong, that works too. I tend to use the partition in three summands because that reduces proving the joint continuity to proving the separate continuity in both arguments plus the joint continuity in the one point $(0,0)$. That simplifies things considerably when proving continuity of bilinear maps on non-normable topological vector spaces. For normable spaces like $\mathbb{R}$, it's not a big difference. – Daniel Fischer Dec 20 '13 at 08:31
  • If it is not too much work I'd be very thankful if you could say more about what is special about $(0,0)$ and also, what role the norms play. – blue Dec 20 '13 at 08:58
  • I now tried something else with my inequality. If $|y-y_0|<{\varepsilon \over 2}$ then $|y|<{\varepsilon \over 2}+|y_0|$. If $\delta = \min ({\varepsilon \over 2}, {\varepsilon \over 2|x_0|}, {\varepsilon \over 2({\varepsilon \over 2} + |y_0|)})$ then $$ |y||x-x_0| + |x_0||y-y_0| < \varepsilon$$ – blue Dec 20 '13 at 09:04
  • @blue The special thing about $0$ is that it's $0$ ;) $0$ times anything is $0$. In normed spaces, you have basically only one neighbourhood, a ball, the rest is scaling, and you have a measure how big something is. In other spaces, you can have different types of neighbourhoods, and no uniform measure of bigness, and then it can be simpler to show 1. "small × small is small", 2. "anything fixed × small is small", 3. "small × anything fixed is small" than to show in one go "(something_1 + something small)×(something_2 + something else small) - something_1 × something_2 is small". – Daniel Fischer Dec 20 '13 at 09:56