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Let $e_1,\ldots,e_j$ be a basis for a finite dimensional normed vector space $X$. I wish to show that the map $(a_1,\ldots,a_n) \mapsto \sum_1^n a_j e_j$ is continuous, where $(a_1,\ldots,a_n)$ has the Euclidean metric and we consider the metric induced by the $1$-norm on $\sum_1^n a_j e_j$.

Let $x = \sum_1^n a_j e_j$ and $y = \sum_1^n b_j e_j$. I can observe the following:

$$\begin{align*} \|x-y\|_2 &= \|\sum(a_j-b_j)e_j\| \\ &\le \sum \|(a_j-b_j)e_j\|_2 \\ &= \sum |a_j-b_j| \|e_j\|_2 \\ &\le \sum |a_j-b_j| \max_{1\le j\le n} \|e_j\|_2 \\ &= \|x-y\|_1 \max_{1\le j\le n} \|e_j\|_2. \end{align*}$$

Now, for any $\epsilon$, I set $\delta = \frac{\epsilon}{\max_{1\le j \le n} \|e_j\|_2}$.

But I am having trouble seeing how $\|x-y\|_2 < \delta$ gets me $\|x-y\|_1 < \epsilon$ in this case.

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  • $\begingroup$ Just show that is continuous in $0$, you know why this is enough? $\endgroup$ – Luis Felipe May 7 '15 at 4:15
  • $\begingroup$ Yes, because it is a bounded linear functional. I suppose that is easier. $\endgroup$ – Emily May 7 '15 at 4:20
  • $\begingroup$ For continuity of your function, call $T$, you must show $\|x-y\|_2<\epsilon$, whenever $\|x-y\|_1<\delta$ and you did it. But to prove continuity of $T^{-1}$ you should apply some other trick ! Work on unit sphere of $\mathbb{R}^n$. $\endgroup$ – Fardad Pouran May 7 '15 at 4:42
  • $\begingroup$ @fardadpouran can you elaborate in an answer? I don't need a full answer, just a nudge. $\endgroup$ – Emily May 7 '15 at 12:18
  • $\begingroup$ Ah, perhaps I have it. $\endgroup$ – Emily May 7 '15 at 13:49
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$\newcommand{\id}{\operatorname{id}}$First, $X$ is "exactly" same as $\mathbb{R}^n$ w.r.t euclidean norm.

Let $\id$ be the transformation $(a_1,\cdots,a_n)\mapsto \sum a_je_j$. What you wrote in your question, proves the continuity of $\id$. To prove continuity of $\id^{-1}$, it's sufficient to prove that the set $\left\{\frac{\|x\|_e}{\|x\|_2}\;;x\neq0\right\}$ is bounded. Suppose $S^{n-1}$ is the unit sphere of $\mathbb{R}^n$ w.r.t euclidean norm.

There are several approaches to show this, but all are common in using compactness of the unit Sphere.

One way is this :

$\id(S^{n-1})$ is compact in $X$. Since $0\notin S^{n-1}$, therefor $0\notin \id(S^{n-1})$. It means that $0$ is the exterior point of $\id(S^{n-1})$, i.e there is $\delta>0$ such that $B_\delta(0)\cap \id(S^{n-1})=\emptyset$.

Therefore, for any $x\in X$, if $\|x\|_2<\delta$, then $\|x\|_e\neq1$. I'll prove that $\|x\|_e<1$ later (if you ask). Since $\left\{\frac{\|x\|_e}{\|x\|_2}\;;x\neq0\right\}=\left\{\|x\|_e\;;\;\|x\|_2=\frac\delta2\right\}$, so the set is bounded by $\frac2\delta$. $\qquad\qquad\square$

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    $\begingroup$ This is precisely the construction I was going for -- proof of equivalence of norms. I just got caught up in a morass of details, as you could see. Many thanks for clearing it up for me :) $\endgroup$ – Emily May 7 '15 at 15:39
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Ok, I've made a real mess of this, but I've figured it out. Simple is good.

Let $T : (a_1,\ldots, a_n) \to \sum a_j e_j$. It is obvious that $T$ is linear. We explore its continuity at zero.

Let $\|\sum a_je_j\|_1 = \sum |a_j| < \epsilon$ for any $\epsilon > 0$.

Then, $\left(\sum |a_j|\right)^2 < \epsilon^2$, and hence $\left(\sum a_j^2\right)-2\sum_{j=1}^n\sum_{k=j+1}^n |a_j a_k| < \epsilon^2$. Let $$\delta = \sqrt{\epsilon^2+2\sum_{j=1}^n\sum_{k=j+1}^n |a_j a_k|}$$ and we have $\|(a_1,\ldots,a_n)\|_2 = \sqrt{\sum a_j^2} < \delta$ when $\sum |a_j| < \epsilon$.

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  • $\begingroup$ What was your goal to prove ? In your question text, you've completely proven that $(a_1,\cdots,a_n)\mapsto \sum a_je_j$ is Lipschitz and hence continuous ! $\endgroup$ – Fardad Pouran May 7 '15 at 15:15
  • $\begingroup$ @FardadPouran I was trying to wrestle it into $\epsilon$-$\delta$ form. But you are right about the Lipschitz criterion -- I had forgotten all about that. $\endgroup$ – Emily May 7 '15 at 15:39

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