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A convex set $M$ is called convex body if it has nonempty interior. Interior $I(M)$ consists of elements of $M$ such that $x+ty$ is in $M$ for any $y$ and positive number $r=r(y)$ such that absolute value of $t$ smaller than $r$.

In $\ell^2$, $M$ is the set of sequences such that infinite sum of squares less than or equal to $1$. Show that $M$ is convex but not convex body.

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    You have to specify what vector space you're working in. If it's $\ell_2$, then $M$ is a convex body. If it's some space $V$ that properly contains $\ell_2$, try taking $y$ to be a member of $V$ that is not in $\ell_2$.2012-01-03

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Hint: To show that the set of sequences with $l^2$ norm less than or equal to $1$ is convex, let $x_n,n\in \mathbb{N}$ and $y_n,n\in \mathbb{N}$ be sequences with $l^2$ norm less than or equal to $1$ and observe that $(tx_i+(1-t)y_i)^2 \leq tx_i^2+(1-t)y_i^2$ for $t\in [0,1]$, hence the $l^2$ norm of the sequence $x+y$ is at most $t\|x\|_2+(1-t)\|y\|_2$.

As Robert said in his comment, this set is a convex body as a subset of $l^2$, so you need to tell us what vector space you're working over. If forced to guess, I would say you're probably working over $l^\infty$, as any square-summable sequence in $l^\infty$ plus a sequence $\epsilon,\epsilon,\ldots\in l^\infty$, which has $l^\infty$ norm $\epsilon$, is not square-summable. But again, you could clear this up if you tell us what vector space you're working over.

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    our space is l2. but i did not not defined M correctly. M is the set of the sequences whose infinite sum of n^2(x_n)^2 is less than or equal to 1. in showing interior points, y should be taken arbitrarily in l2. thank you2012-01-04