# Proof that every normed vector space is a topological vector space

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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I do not like the wording of this question. A topology is a structure you equip onto a set, not a property that a set can have. You should say that the norm induces a topology making $V$ a topological vector space. –  Qiaochu Yuan Jul 7 '12 at 16:24
@QiaochuYuan The first sentence? Or where? –  Rudy the Reindeer Jul 7 '12 at 16:42
Yes, the first sentence (equivalently the title). I should mention that this is a minor nitpick; I just think most people use the word "is" too loosely. –  Qiaochu Yuan Jul 7 '12 at 16:51
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. –  wildildildlife Jul 7 '12 at 17:51

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 $|v-v_0|\leq \delta$.
Sorry, how did you get $$|\alpha_0-\alpha|\| v_0\|+|\alpha|\| v-v_0\| \leq |\alpha_0-\alpha|(\| v_0\|+\| v-v_0\|)+|\alpha_{\color{\red}{0}}|\| v-v_0\|$$? I can see that $$|\alpha_0-\alpha|\lVert v_0\rVert+|\alpha|\lVert v-v_0\rVert \leq |\alpha_0-\alpha|(\lVert v_0\rVert+\lVert v-v_0\rVert)+|\alpha|\lVert v-v_0\rVert$$ –  Rudy the Reindeer Jul 7 '12 at 16:38
I wrote $|\alpha|\leq |\alpha-\alpha_0|+|\alpha_0|$. –  Davide Giraudo Jul 7 '12 at 16:43