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The topology induced by the norm of a normed vector space is such that the space is a topological vector space.

Can you tell me if my proof is correct? Of course we have to show that addition and scalar multiplication are continuous with respect to the product topology (induced by the norm).

(1) To show that $(x,y) \mapsto x + y$ is continuous let $\varepsilon > 0$. I can show that the norm $\|\cdot\|_{V \times V}: V \times V \to \mathbb R$ defined as $\|(x,y) - (x_0, y_0) \| = \|x-x_0\| + \|y_0 - y\|$ induces the same topology as the product topology on $V \times V$. Hence we can choose $\delta = \varepsilon$ to get $ \| (x+y) - (x_0+y_0)\| \leq \|x-x_0\| + \|y-y_0\| < \delta = \varepsilon$

(2) To show that $V \times K \to V$, $(v, \alpha) \mapsto \alpha v$ is continuous at $(v,\alpha)$, observe that $\| \alpha v - \beta w\| = \| \alpha v - \beta w + \alpha w - \alpha w\| = \|\alpha(v-w) + (\alpha - \beta) w\| \leq |\alpha| \|v-w\| + |\alpha - \beta| \|w\|$

Hence $\| \alpha v - \beta w\| < \varepsilon$ if $\|v-w\| < \frac{\varepsilon}{2 |\alpha|}$ and $|\alpha - \beta| < \frac{\varepsilon}{2 \|w\|}$. Unfortunately, the second inequality depends on $w$. How do I make it independent of $w$? Thanks.

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    Hint: recall (from your introductory analysis course) the proof of the sum and product rule for limits in $\mathbb{R}$. Then put norm signs in appropriate places.2012-07-07

2 Answers 2

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The first point is fine. For the second, fix $(v_0,\alpha_0)\in V\times K$ and $\varepsilon >0$. We have to find $\delta>0$ such that if $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$ then $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$. We have \begin{align} \lVert \alpha_0v_0-\alpha v\rVert&\leq \lVert \alpha_0v_0-\alpha v_0\rVert+ \lVert \alpha v_0-\alpha v\rVert\\ &=|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert\\ &\leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha_0|\lVert v-v_0\rVert. \end{align} We take $\delta$ such that $\delta^2+\delta(\lVert v_0\rVert+|\alpha_0|)\leq \varepsilon$ (which is possible).

In this case, $\lVert \alpha_0v_0-\alpha v\rVert\leq \varepsilon$ when $|\alpha-\alpha_0|\leq \delta$ and $|v-v_0|\leq \delta$.

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    I wrote $|\alpha|\leq |\alpha-\alpha_0|+|\alpha_0|$.2012-07-07
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Let fix $\alpha$ and $ x$ such that,

$\|x-x_0\|<1 $ then, $\|x\| \le \|x-x_0\|+\|x_0\|\le \|x_0\|+1$

\begin{align}\|\alpha x -\alpha_0x_0\| &= \|\alpha x -\alpha_0 x+\alpha_0 x-\alpha_0x_0\| \\&\le \|x\||\alpha -\alpha_0| +|\alpha_0| \|x-x_0\|\\&< (\|x_0\|+1)|\alpha -\alpha_0| +(|\alpha_0|+1) \|x-x_0\|\\&\le \max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\left[|\alpha -\alpha_0| + \|x-x_0\|\right]}\\&\le 2\max\left[(\|x_0\|+1),(|\alpha_0|+1)\right]\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right]} \end{align}

for any $\varepsilon>0$ if you take $\color{red}{\delta= \min\left(1, \frac{\varepsilon}{ 2\max(\|x_0\|+1),(|\alpha_0|+1)}\right)}$ Then for any $x$ and $\lambda$ such that, $\color{red}{\max\left[|\alpha -\alpha_0| , \|x-x_0\|\right] <\delta}$ you get, $\color{blue}{ \|\alpha x -\alpha_0x_0\|<\varepsilon}$

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    @JackD'Aurizio it is up to you delete what you want but. actually I discover that the other post was duplicate after somebody rise up that up. in a meanwhile I had already answered the question. may be you got back and read the comments on that post . and check the timing. you will see that it was not intentional. I don't fall in to the same trap twice2017-11-17