# Thread: Question in Functional Analysis

2. ## Re: Question in Functional Analysis

May i know what int E means ?

3. ## Re: Question in Functional Analysis

I think it is interior of E.

4. ## Re: Question in Functional Analysis

interior of E denoted ( int(E) or A^o

5. ## Re: Question in Functional Analysis

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 $\displaystyle \left\{ {x,y} \right\} \subset \overline{E} \;\& \;\varepsilon > 0$ 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 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.