Hello,

I need to prove the following but I am struggling to see where to start:

The image of a convex set under an affine transformation i.e. the set for , where

However, I know how to prove convex.

Thanks in advance,

Best regards,

bravegag

Printable View

- Feb 28th 2010, 01:38 AMbravegagprove convexity
Hello,

I need to prove the following but I am struggling to see where to start:

The image of a convex set under an affine transformation i.e. the set for , where

However, I know how to prove convex.

Thanks in advance,

Best regards,

bravegag - Feb 28th 2010, 04:32 AMDinkydoe
The important part is proving that: convex convex. Trivially, the set is convex as well.

S is convex, meaning for all and any we have .

Now, for you to prove that is convex, prove that any convex combination . - Feb 28th 2010, 05:03 AMbravegag
Hello Dinkydoe,

Thank you for your answer! I ended with this below as proof ... does it make sense to you?

Proof: . Now to prove that is convex, I need to prove that any convex combination

Let and . Then applying the Convex definition:

substituting

Finally by convex definition where convex and then is convex too.

I am not a mathematician but interested to learn the course Convex Optimization ... a bit too many proofs for a poor programmer like me (Worried)

TIA,

Best regards,

Giovanni - Feb 28th 2010, 03:43 PMDinkydoe
Your proof is indeed correct:

I think the conclusion you gave:

Quote:

Finally by convex definition http://www.mathhelpforum.com/math-he...e31ecc24-1.gif where http://www.mathhelpforum.com/math-he...fa3440e5-1.gif convex and http://www.mathhelpforum.com/math-he...ce353ae6-1.gif then http://www.mathhelpforum.com/math-he...a852cd88-1.gif is convex too.

For any we have for any

Since that's what you've shown, and I think that's what you mean ;)