compact convex problem Would you please help me in how to show that:
If $T : X \to Y$ is a continuous linear map between two locally convex spaces and $K \subset X$ is compact convex, then:


*

*The continuous image of compact convex space is a compact convex. 

*If $y$ ia an extreme point of $T(K)$,  then $\exists$  $x \in X$ an extreme in $K$ such that $Tx =y$
Thank you,
Saba
 A: The first point is obvious:


*

*Compactness is preserved by continuous maps so $T(K)$ is compact.

*Convexity is preserved by linear maps, so $T(K)$ is convex.



As for the second point, we need to know a bit more, I think.
First of all, I assume that your definition of locally convex includes Hausdorff, otherwise there are silly counterexamples (see right at the end of the answer).
The idea is the following: If $y \in T(K)$ is extremal then $F = \{x \in K\,:\,T(x) = y\}$ is a face of $K$ and by the Kreĭn-Mil'mans theorem every face contains an extremal point. If you know what a face is and just want to see see why $F$ is a face, you can ignore the next section and jump right to the claim afterwards.
Also, I overlooked the homework tag, so maybe I shouldn't have given all away below, stop reading now if you want to think for yourself.

Let me elaborate on this:
Recall that a face $F$ of a convex set $K$ is a non-empty and closed subset of $K$ with the following property: If $x,y \in K$ are any points for which there is $0 \lt \alpha \lt 1$ such that $(1-\alpha)x + \alpha y \in F$ then we must have $x,y \in F$ already. (Think of the faces of a cube, for example).
Let me sketch the proof of Kreĭn-Mil'man's theorem which states: 

If $K$ is a compact convex set in a locally convex space, then it is the closed convex hulls of its extremal points.

Proof:


*

*For $x \in K$ the set $\{x\}$ is a face if and only if $x$ is extremal.

*If $\phi: X \to \mathbb{R}$ is a continuous linear functional and $F \subset K$ is a face, then $\{x \in F\,:\,\phi(x) \text{ is minimal}\}$ is a face again.

*Order the faces of $K$ by reverse inclusion. The intersection of a chain of faces is a face, thus every face contains a minimal face by Zorn's lemma. 

*Combine 2. and Hahn-Banach to see that a minimal face cannot contain more than one point.


Taking 3. and 4. together we see that we have proved what we need for solving the exercise.

Every face of a compact convex set contains an extremal point.

Since it would be a pity to stop here, let me finish the argument for Kreĭn-Mil'man:
Since $K$ is a face of itself, there is at least one extremal point. If the closed convex hull $C$ of the extremal points were not all of $K$ then there were a point $z$ and a continuous linar functional $\phi$ such that $\phi(z) \lt \phi(c)$ for all $c \in C$. But by applying 2. to $\phi$ we find a face disjoint from all extremal points, a contradiction.


Claim: If $y \in T(K)$ is extremal, then $F = \{x \in K\,:\,T(x) = y\}$ is a face.

In particular, $F$ contains an extremal point by what I just said.
Proof of the claim: Obviously, $F$ is closed and non-empty. If $F$ were not a face, there would be $x_0,x_1 \in K \smallsetminus F$ and $0 \lt \alpha \lt 1$ such that $(1-\alpha)x_{0} + \alpha x_1 \in F$, so $(1-\alpha) T(x_0) + \alpha T(x_1) = y$, but $T(x_{0}) \neq y \neq T(x_1)$, so $y$ is not extremal, a contradiction.

To see that we need to assume at least that $X$ is Hausdorff, let $X = \mathbb{R}^2$ with the semi-norm $|(x,y)| = |x|$ and let $Y = \mathbb{R}$ with the usual topology. Let $T: X \to \mathbb{R}$ be projection on the first coordinate, $T(x,y) = x$. This map is continuous and obviously the $y$-axis $K$ in $X$ is compact and convex and contains no extremal point. However, $T(K) = \{0\}$ and the point $\{0\}$ is extremal. By construction there is no extremal point in $K$ contradicting the exercise.
A: I think you will find this easier if you recall that "convex" can be defined in terms of inequalities and that the linear map will preserve those features.
