# Show the $l^2$ ellipsoid is close : $\left\{(x_n) \in l^2 : \sum\frac{x_n^2}{a_n^2} \le 1\right\}, a_n \to 0$

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 $l^2$.

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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

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 $S$.

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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