Re: Convex bodies Shadow.

Well for

For

a) Any element of S(u;X) has the form $\displaystyle ut + x \text{ where } t >0 \text{ and } x \in X $ Well, to show convexity you have to show that for any $\displaystyle p \in [0,1] $ $\displaystyle p(tu + x) + (1-p)(tu + x) \in S(u;X) $ So multiplying the p through we get $\displaystyle ptu + px + (1-p)tu + (1-p)x $ since X is convex, $\displaystyle px + (1-p)x = j \in X $. So we got $\displaystyle ptu + (1-p)tu + j $, then factor out the t to get $\displaystyle (pu+(1-p)u)t + j $, which is $\displaystyle (pu + u - pu)t + j $ which is $\displaystyle ut + j \in S(u; X) $ since $\displaystyle j \in X $

b)

Assume X is open.

To show that S(u;X) is open, for any element $\displaystyle j \in S(u;X) $ i need to show that there exists a positive radius r such that this implication holds, if $\displaystyle p \in B_r(j)$ then p must be in S(u;X).

Now, since $\displaystyle j \in S(u;X) $ j is of the form $\displaystyle j = tu + x $ where $\displaystyle x \in X $. Since X is open, there exists a radius $\displaystyle r_1 > 0 $ such that this implication holds, if $\displaystyle \bar{x} \in B_{r_1}(x)$ then $\displaystyle \bar{x} \in X $.

Now i will show that for the radius $\displaystyle r_1 > 0 $ derived from j = tu + x (it is the radius for x, an element of X ), if $\displaystyle p \in B_{r_1}(j) $ then $\displaystyle p \in S(u;X) $.

Now, assume $\displaystyle p \in B_{r_1}(j) $ So $\displaystyle ||p -(ut + x) || = ||p - ut - x || < r_1 $. So this means that $\displaystyle ||(p-ut) - x || < r_1 $ which means $\displaystyle p - ut \in X $ Thus $\displaystyle ut + (p-ut) \in S(u;X) $, thus $\displaystyle p \in S(u; X) $ which means that S(u;X) is open.

Re: Convex bodies Shadow.

thanks for the help jakncoke !!!!!!

Re: Convex bodies Shadow.

I re open the thread, cause im missing the inclusion X in S(u,X) knowing that X is open. I have no clue.

Thanks!