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- Dec 1st 2012, 12:02 PMsoso123Question in Functional Analysis
- Dec 1st 2012, 01:48 PMjakncokeRe: Question in Functional Analysis
May i know what int E means ?

- Dec 1st 2012, 01:53 PMcoolgeRe: Question in Functional Analysis
I think it is interior of E.

- Dec 1st 2012, 02:23 PMsoso123Re: Question in Functional Analysis
interior of E denoted ( int(E) or A^o

- Dec 1st 2012, 03:33 PMPlatoRe: Question in Functional Analysis
There are many, many different notations for the interior of a set. That is why you should always define the terms you post.

Now to question. The second is easier to prove that the first.

If $\displaystyle \left\{ {x,y} \right\} \subset \overline{E} \;\& \;\varepsilon > 0$ then suppose that $\displaystyle \alpha~\&~\beta$ are positive numbers such that $\displaystyle \alpha+\beta=1$. Now $\displaystyle (\exists x_1\&y_1)\in E$ such that $\displaystyle \|x-x_1\|<\varepsilon~\&~\|y-y_1\|<\varepsilon$

Now you need to show that $\displaystyle \|\alpha x+\beta y-(\alpha x_1+\beta y_1)\|<\varepsilon$ which shows that $\displaystyle (\alpha x+\beta y)\in\overline{E}$ WHY?

The first problem is a bit more involved. You must show that no point on the path from two points in $\displaystyle \text{INT}(E)$ contains a point of the $\displaystyle \beta(E)$, the boundary of $\displaystyle E$ (see what I mean about defining terms?)

There is a standard concept of anof a convex set. At first, that is what I thought you meant. However, that does not appear to be the case here. So you most lookup that concept. Then prove the theorem that:*internal point**each point of the interior of a convex subset of a topological vector space is an internal point.*

The proof depends upon the continuity of scalar multiplication.

Be warned: I will not do this for you.