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Let $X$ be a normed space over scalar field $\mathbb{K}$. I have to show that scalar multiplication map is continuous.

So take $X\supset(x_n)\rightarrow x$ and $\mathbb{K}\supset(k_n)\rightarrow k$. I have to show that $k_n x_n \rightarrow kx$

$$\| k_n x_n - kx\|=\| k_n x_n - kx_n+k x_n- kx\|\le\| k_n x_n - kx_n \|+ \|k x_n- kx\|\le$$ $$\le |k_n-k|\|x_n\|+|k|\|x_n-x\| \longrightarrow 0$$

Is that ok? I've seen some proofs that are a lot longer and use $\delta$, $\epsilon$ notation. Is that really necessary? I don't see any flaws in my proof.

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    $\begingroup$ It is not necessary, but the $\epsilon$-$\delta$ proof is not much longer either. You need to show that $\|x_n\|$ is bounded. $\endgroup$ – copper.hat Jun 13 '14 at 14:54
  • $\begingroup$ Norm is continuous so it is bounded, yes? $\endgroup$ – luka5z Jun 13 '14 at 15:40
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    $\begingroup$ Yes, if $x_n \to x$ then $\|x_n\| \to \|x\|$ and and convergent sequence of reals is bounded. My point was that an $\epsilon$-$\delta$ proof is not much longer if you add the details... $\endgroup$ – copper.hat Jun 13 '14 at 15:57
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Define $ k : X \rightarrow X,\ k(x)=kx$

If $\parallel x-y\parallel<\delta$, then $$ \parallel kx-ky\parallel = k\parallel x-y\parallel < k \delta < \epsilon $$

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    $\begingroup$ Scalar multiplication is $\cdot: \mathbb{K} \times X \to X$, not $X \to X$. $\endgroup$ – copper.hat Jun 13 '14 at 14:56
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    $\begingroup$ Use $\|(k,x)\| = |k|+\|x\|$ for example. $\endgroup$ – copper.hat Jun 13 '14 at 15:07
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    $\begingroup$ If $|k-l|+\parallel x-y\parallel =\parallel (k,x)-(l,y)\parallel < \delta$, then $\parallel kx-ly \parallel \leq \parallel kx-ky \parallel +\parallel ky-ly \parallel\leq k\delta +\delta \parallel y \parallel \leq k\delta +\delta ( \delta + \parallel x \parallel ) $ $\endgroup$ – HK Lee Jun 13 '14 at 15:18
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    $\begingroup$ Sorry, I haven't had my caffiene yet. The norm I had above (which I will delete) makes no sense. I originally wrote 'Use the product topology', then it seemed cuter to use the formula I wrote. $\endgroup$ – copper.hat Jun 13 '14 at 15:21
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    $\begingroup$ @HKLee It seems that using the norm for the product topology would be correct here: your answer only proves continuity in the vector argument. One can easily prove continuity in the scalar argument by the property of the norm ( $\parallel kx\parallel = |k|. \parallel x\parallel$) but even then separate continuity in the two arguments doesn't prove total continuity. I'm happier with your comment, concluding ... $\parallel kx-ly \parallel \leq |k|\delta +\delta ( \delta + \parallel x \parallel )$ let $\delta \lt min(1, \epsilon / (|k| + 1 + \parallel x \parallel))$ ..... $\endgroup$ – Tom Collinge Mar 15 '17 at 9:31

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