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.

  • $\begingroup$ 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

$\newcommand{\id}{\operatorname{id}}$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 $\frac2\delta$. $\qquad\qquad\square$

  • 1
    $\begingroup$ 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

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 $\sum |a_j| < \epsilon$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.