1
$\begingroup$

A (real) linear topological space is a real linear space (vector space) $Ε$ with a Hausdorff topology such that:

I) vector addition is continuous

II) scalar multiplication is continuous

For $x$ and $у$ in $E$, denote by $L(x, y)$ the set of all points $z$ such that $z= λ_1x +λ_2y$ with $0 \le λ_i \le 1$ and $λ_1 + λ_2 = 1$.

A subset $A$ of $E$ is convex iff whenever $x$ and $y$ belong to $A$,then $L(x,y) \subset A$

My Question:-

If $A$ is convex and $x \in A^{\mathrm{o}} $, $у \in \bar{A}$, show that $L(x, у)-{у} \subset А$

I Know that $A^{\mathrm{o}} $ and $\bar{A}$ are convex sets, but what should I do then ??

$\endgroup$
  • $\begingroup$ Draw a little picture of the situation. If $x \in A^\circ$ then there is some open $U\subset A$ containing $x$. Now shift and scale (shrink) the set $U$ as you move from $x$ to $y$ 'linearly' so that the resulting set stays inside $A$. $\endgroup$ – copper.hat Oct 31 '17 at 2:54
  • $\begingroup$ @copper.hat: that picture works well to show that $x\in \mathring{A}$, $y\in A$ implies $[x,y)\subset \mathring{A}$. Here $y \in \bar{A}$. $\endgroup$ – Orest Bucicovschi Oct 31 '17 at 3:16
  • $\begingroup$ @orangeskid: It works for $y$ in the closure as well. $\endgroup$ – copper.hat Oct 31 '17 at 3:20
  • $\begingroup$ In my opinion, this particular fact captures the essence of convexity. $\endgroup$ – copper.hat Oct 31 '17 at 3:21
  • 1
    $\begingroup$ @copper.hat: I agree, the cone without vertex lies inside. $\mathring{A}$. Good call, also about the relative interior. $\endgroup$ – Orest Bucicovschi Oct 31 '17 at 3:33
1
$\begingroup$

HINT:

In fact you can show that if $x\in \mathring{A}$, and $y\in \bar A$, then $[x,y)\in \mathring{A}$. Let $\lambda \in (0, 1]$. We want to show that $$\lambda x + (1-\lambda)y + \delta_{\lambda} \in A$$ if $\delta_{\lambda}$ is small enough. What we know is that in every neighborhood of $0$ there exists $\delta$ so that $y+\delta \in A$. So now we rewrite $$\lambda x + (1-\lambda)y + \delta_{\lambda}= (1-\lambda)(y + \delta) + \lambda( x- \frac{1-\lambda}{\lambda}\delta + \frac{1}{\lambda}{\delta_\lambda})$$ Recall that $\lambda \in (0,1]$ is fixed. If $\delta$ and $\delta_{\lambda}$ are small enough, $- \frac{1-\lambda}{\lambda}\delta + \frac{1}{\lambda}{\delta_\lambda}$ will lie in the neighborhood $W_{x}$ with the property that $x+W_{x}\subset A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.