Showing a set is convex, but not a convex body

Question

I am stuck on the following homework problem:

Let M be the set of all points in $l_{2}$ satisfying $\Sigma_{n=1}^{\infty} n^{2}x_{n}^{2} \leq 1$. Prove that M is a convex set, but not a convex body.

I have proved it is a convex set, but proving it is not a convex body has stumped me. I have the following:

Let x,y$ \in$ M, let |t|$\lt \epsilon$ for some positive epsilon. Then for any fixed t such that |t|$\lt \epsilon$, we have $\Sigma_{n=1}^{\infty} n^{2}(x_{n}+ty_{n})^{2} \leq 1 + t^{2} + 2t$ (using Cauchy-Schwarz inequality). $1 + t^{2} + 2t \leq 1$ iff $t \leq -2$.

Now I'm unsure of where to go from here. I know that this means that if t is not less than -2, the segment is not in M and hence M is not a convex body. But all this seems to say to me is that I simply need to find an epsilon larger than 2.

Answer 1

The function $x \mapsto x_n^2$ is convex an non negative, hence $f: l_2 \to [0,\infty]$ given by $f(x) = \sum_n n^2 x_n^2$ is convex, and hence $f^{-1} ((-\infty,1])$ is convex.

Let $h_n = {1 \over n}$, then $h \in l_2$.

Suppose $x\in M$, then $\sum_n n^2 (x_n+t h_n)^2 \ge t^2 \sum_n n^2 h_n^2$, and hence $x+t h \notin M$ for any $t \neq 0$.

In particular, $M$ has an empty interior.

Answer 2

Suppose $B(a,r)\subset M.$ Then

$$\tag 1 v_m=a+\frac{r}{2}e_m \in M$$

for every $m.$ But the square of the $m$th component of $v_m$ is $a_m^2 + a_mr + r^2/4.$ Thus

$$ \tag 2 m^2(a_m^2 + a_mr + r^2/4)\le 1$$

for all $m.$ Since $a_m\to 0,$ the expression in the parentheses $\to r^2/4,$ hence $(2)\to \infty,$ contradiction.