Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show the $l^2$ ellipsoid $$ S = \left\{(x_n) \in l^2 : \sum\frac{x_n^2}{a_n^2} \le 1\right\}, $$ is closed, with $(a_n)$ a sequence of positive numbers such that $$ \lim_{n \to \infty} a_n =0. $$

I tried considering a Cauchy sequence in $S$, and an open ball in the complement $\tilde S=l^2\setminus S$, but neither worked.

Let $\{x(k)\}$ be a sequence in $S$, as $l^2$ is complete, $x(k)=(x_n (k))_n$ goes to some $x=(x_n)$ in $l^2$. Then how to show $$ \sum_n \frac{x_n^2}{a_n^2} \le 1? $$

Let $x=(x_n)$ be a point in $\tilde S$, then $$ \sum \frac{x_n^2}{a_n^2} > 1. $$ Now, there should be a $\delta > 0$ for which whenever $y \in B\left(x,\delta \right)$ $$ \sum \frac{y_n^2}{a_n^2} > 1. $$

I know $(a_n)$ is bounded as it is convergent, and $S$ is also bounded in $l^2$.

share|cite|improve this question
Show that for each $m$, $\sum\limits_{n=1}^m {x_n^2\over a_n^2}\le 1$. Then take the limit ae $m\rightarrow\infty$. – David Mitra Feb 17 '13 at 14:07
up vote 1 down vote accepted

Previous answer was wrong, now expanding on David Mitra's comment.

Let $(x^{(k)})$ be a sequence in $S$ and assume it converges to $x$ in $\ell^2$.

You have to prove that $x$ belongs to $S$.

First note that $(x^{(k)})$ converges afortiori pointwise to $x$.

Fix $m$.

$$ 0\leq \sum_{n=1}^m\frac{(x_n^{(k)})^2}{a_n^2} \leq \sum_{n=1}^{+\infty}\frac{(x_n^{(k)})^2}{a_n^2}\leq 1 $$ for all $k$.

Letting $k$ tend to $+\infty$, we get $$ 0\leq \sum_{n=1}^m\frac{x_n^2}{a_n^2}\leq 1. $$

Now this is true for every $m$, so $$ 0\leq \sum_{n=1}^{+\infty}\frac{x_n^2}{a_n^2}\leq 1. $$

So $x$ belongs to $S$.

share|cite|improve this answer
I don't see the dominating function which permits application of the DCT. I see that $\sum_n \frac{\left(x_n^{(k)}\right)^2}{a_n^2} \le 1$, but $\sum_n 1 = \infty$. – Nicolas Essis-Breton Feb 17 '13 at 15:08
@NicolasEssis-Breton Oh yes, you're right, sorry about that. – 1015 Feb 17 '13 at 15:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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