- #1

Oster

- 85

- 0

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).

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In summary, To prove that all open balls in a Normed Linear Space are convex, we can use the definition of convexity and rewrite the point z as (1-t)x + ty. By using the norm definition, we can show that ||z|| is less than the radius of the ball, thus proving that z belongs to the ball. Finally, we need to use the correct terminology and rewrite the original points as p - a and q - a in order to complete the proof.

- #1

Oster

- 85

- 0

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).

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- #2

tiny-tim

Science Advisor

Homework Helper

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- 258

Hi Oster!

Hint: rewrite z as (1-t)x + ty …

Hint: rewrite z as (1-t)x + ty …

then what can you say about ||z|| ?

- #3

Oster

- 85

- 0

||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?

- #4

tiny-tim

Science Advisor

Homework Helper

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(have an leq: ≤ )

Oster said: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

- #5

Oster

- 85

- 0

Thanks much!

An NLS, or normed linear space, is a vector space equipped with a norm function that measures the size or length of a vector.

An open ball in an NLS is a set of all points within a certain distance from a given point, where the distance is defined by the norm function.

To prove that open balls in an NLS are convex, we need to show that for any two points in the ball, the line segment connecting them is also contained within the ball. This can be done by using the triangle inequality and properties of the norm function.

It is important to prove that open balls in an NLS are convex because it is a fundamental property of NLS that is used in many proofs and applications. It also allows us to define and study other important concepts like convex sets and convex functions.

No, there are no counterexamples to this statement. The proof for open balls in an NLS being convex holds for any NLS, regardless of its dimension or properties.

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