# Question in Functional Analysis

• Dec 1st 2012, 12:02 PM
soso123
Question in Functional Analysis
• Dec 1st 2012, 01:48 PM
jakncoke
Re: Question in Functional Analysis
May i know what int E means ?
• Dec 1st 2012, 01:53 PM
coolge
Re: Question in Functional Analysis
I think it is interior of E.
• Dec 1st 2012, 02:23 PM
soso123
Re: Question in Functional Analysis
interior of E denoted ( int(E) or A^o
• Dec 1st 2012, 03:33 PM
Plato
Re: Question in Functional Analysis
Quote:

Originally Posted by soso123
interior of E denoted ( int(E) or A^o

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

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