Prove Open Balls in an NLS are convex.

[Oster]

Given a Normed Linear Space, prove that all open balls are convex.

A, a subset of the space is said to be convex if, for all pairs of points (x,y) in a, the point
z = x + t(y-x) belongs to A. (t goes from 0 to 1).

 

[tiny-tim]

Hi Oster!

Hint: rewrite z as (1-t)x + ty …

then what can you say about ||z|| ? ​

 

[Oster]

Hmm

||z|| = ||(1-t)x + ty|| <= (1-t)||x|| + t||y|| [norm definition]

But, if I'm working with a ball centered at 0 with radius r then

||x|| and ||y|| would be less than r implying that

||z|| < (1-t)r + tr = r

Which implies z belongs to this ball around 0 with radius r.

So, all that's left to prove is that any ball B(a,r) = a + B(0,r)?

Am I correct?

 

[tiny-tim]

Hi Oster!

(have an leq: ≤ )

Hmm

||z|| = ||(1-t)x + ty|| <= (1-t)||x|| + t||y|| [norm definition]

But, if I'm working with a ball centered at 0 with radius r then

||x|| and ||y|| would be less than r implying that

||z|| < (1-t)r + tr = r

Which implies z belongs to this ball around 0 with radius r.

That's right!

So, all that's left to prove is that any ball B(a,r) = a + B(0,r)?

you started with the wrong terminology …

if your two original points are p and q in a ball of centre a, then your x is p - a, and your y is q - a ​

 

[Oster]

Thanks much!