# Need some pointers

1. Mar 12, 2005

### Oxymoron

I am working through some exercises and need some help on a couple of questions.

1. Show that in any inner product space $$X$$ that

$$B[x;r] = x+rB[X] := \{x+ry\,:\, y\in B[X]\}$$

where $$B[X]$$ is the closed unit ball.

2. Show that the closed unit ball is convex.

I have thought about these questions, and I can picture them. However, I cant begin to prove them. Can anyone get me started? I only need pointers.

Thanks.

2. Mar 12, 2005

### Hurkyl

Staff Emeritus
Quite often, a good place to start is with the definitions.

3. Mar 12, 2005

### Oxymoron

Hello Hurkyl, I'm just going to show you what I've done for 2.

The definition for a convex subset C is that the line segment between any two points of the subset entirely lies within C. That is, a subset shaped likean annulus or a heart is not convex. The unit ball $$B_X$$, as it is defined, is a circle in $$\mathbb{R}^2$$, or a sphere in $$\mathbb{R}^3$$, etc... So intuitively it is going to be convex.

Another way of writing the definition of a convex subset is

$$\forall \, x,y \in C$$ and $$\forall \lambda\in [0,1]$$ we have

$$\lambda x + (1-\lambda) y \in C$$

But I cant find anywhere a procedure on how to show that a subset is convex.??

4. Mar 12, 2005

### Data

The definition of the closed unit ball in an inner product space $$X$$ is $$B\left[X\right]=\{x \in X : |x| \leq 1\}$$. See if that helps~

Last edited: Mar 12, 2005
5. Mar 12, 2005

### Oxymoron

So if $$B_X = B[0;1]= \{x\in X\,:\,|x|\leq1\}$$ then we know that the ball is centred on 0, the radius is $$r=1$$ and that $$|x|\leq1$$.

$$proof$$

Take two points $$x,y\in B_X$$ we know that $$|x|\leq1$$ and $$|y|\leq 1$$. Further, $$0 \leq \lambda \leq 1$$.

That is, the distance between $$x$$ and the origin, and $$y$$ and the origin is always going to be less than or equal to 1. If it wasn't then the points are not in C.

The question at hand requires us to prove that

$$\lambda x - (\lambda - 1)y$$

belongs to C? Informally speaking...

$$0 \leq \lambda x \leq 1$$ and

$$-1 \leq(\lambda - 1) \leq 0$$

which clearly implies

$$-1 \leq(\lambda - 1)y \leq 0$$

From this it is easy to see that $$0 \leq \lambda x - (\lambda - 1)y \leq 1$$

That is

$$\lambda x - (\lambda - 1)y \in [0,1] \in C$$

I know this proof is not very solid, But am I close?

6. Mar 13, 2005

### Data

What is $$C$$? And I don't agree with $$(-1 \leq a \leq 0, \; 0 \leq b \leq 1) \Longrightarrow 0 \leq b - a \leq 1$$, or that statements like these even make sense when discussing elements in an arbitrary inner product space. You have also made an error: the theorem you quoted in your earlier post does not look at $$\lambda x - (\lambda - 1)y$$, but at $$\lambda x - (1 - \lambda)y$$ (oops - you didn't make a mistake, I'm just bad at reading!).

You are reasonably close though.

I will give another hint: do you remember the triangle inequality?

In any inner product space $$X, \; \forall x, y \in X, \ \mbox{we have} \ |x + y| \leq |x| + |y|$$

Also, remember that you should make your bounds as strict as possible!

Last edited: Mar 13, 2005
7. Mar 13, 2005

### Data

I will also note that

doesn't make too much sense, unless we're talking about $$\mathbb{R}$$. You can only talk about magnitudes in general inner product spaces (for example, if $$x \in \mathbb{R}^2$$, the statement $$x \in [0, 1]$$ is nonsense, as is $$[0, 1] \subset \mathbb{R}^2$$, but $$|x| \leq 1$$ makes sense).

Last edited: Mar 13, 2005
8. Mar 13, 2005

### Oxymoron

I just want to make it clear, that I actually missed the lecture on this topic, so I have no resources or clues as to how to go about solving such problems. So please bear with me.

Right, I was being too ambiguous with my statements.

Data, so the proof of this is going to require the use of the triangle inequality?

$$proof$$

Take any $$x,y \in B_X$$ where $$B_X$$ is the unit ball of an inner product space. Then $$B_X$$ is convex if for all $$x,y \in B_X$$ and all $$\lambda \in [0,1]$$ we have

$$\lambda x + (1-\lambda)y \in C$$

(dont ask me where I got that other equation from!?!?) From the definition of the unit ball we know

$$\|x\|\leq 1$$ and $$\|y\|\leq 1$$

Now for any $$\lambda$$ such that $$0 \leq\lambda\leq 1$$

$$|\lambda|\|x\|\leq 1$$ and $$|(1-\lambda)|\|y\|\leq 1$$

From the properties of the norm we can absorb the constants

$$\|\lambda x\|\leq 1$$ and $$\|(1-\lambda) y\|\leq 1$$

Ok, Im going to stop now. I dont think Im making any sense. I think Im going to need some more help.

However, I can kind of see how the triangle inequality is going to help, but not sure how to integrate it into the proof.

9. Mar 13, 2005

### Data

You are making perfect sense, you just need to make your bounds a little stricter. Remember that what you need to prove is that

$$(x, y \in B_X, 0\leq \lambda \leq 1) \Longrightarrow \lambda x + (1-\lambda) y \in B_X$$

What does the statement $$\lambda x + (1-\lambda) y \in B_X$$ mean?

Edit: As I have now noted in my previous post, you didn't actually make a mistake regarding the form of the vector you're looking at at all. I imagined it!

Last edited: Mar 13, 2005
10. Mar 13, 2005

### Data

And yes, the clearest way that I see to do it requires the triangle inequality. I'm sure there are other ways, though :)

11. Mar 13, 2005

### Oxymoron

Actually I'm not quite sure what it means. Could you do me the favour.

I still dont know where to go. I cant just say

$$\|\lambda x\| + \|(1-\lambda)y\| \leq 2$$

then by the triangle inequality you have

$$\|\lambda x + (1-\lambda)y\| \leq \|\lambda x\| + \|(1-\lambda)y\| \leq 2$$

Hence

$$\|\lambda x + (1-\lambda)y\| \leq 2$$

12. Mar 13, 2005

### Data

It just means that $$\| \lambda x + (1 - \lambda)y \| \leq 1$$. So all you're trying to prove is: $$( \| x \|, \ \| y\| \leq 1, \ 0 \leq \lambda \leq 1 ) \Longrightarrow \| \lambda x + (1 - \lambda ) y \| \leq 1$$

Alright. If $$|b| \leq 1$$, can I bound $$|ab|$$ by anything in terms of $$a$$?

Edit: Like I said earlier, your current approach is perfect. You just need to make your bounds a little stricter.

Last edited: Mar 13, 2005
13. Mar 13, 2005

### Oxymoron

You can bound $$|ab|$$ by $$a$$ right. Since the length of $$b$$ is at most 1. Then multiplying $$b$$ by $$a$$ you scale $$b$$ by $$a$$. Unless $$a$$ and $$b$$ are vectors, wait...

Can you bound $$|ab|$$ by $$|(a,b)|$$?

That is

$$|ab| = |a||b| \geq |(a,b)|$$

that doesn't work. Hmmm

Last edited: Mar 13, 2005
14. Mar 13, 2005

### Data

You can, by the Cauchy-Schwartz inequality, but that'll give you a lower bound. You need upper bounds. You can (upper) bound $$|ab|$$ by $$|a|$$. That should be enough for you to solve the problem: Just combine your results and this new fact.

If you need more help just keep asking though :)

15. Mar 13, 2005

### Oxymoron

Data, guess what, I think I got it!!!

$$proof$$

Take any $$x,y \in B_X$$ where $$B_X$$ is the unit ball of an inner product space. $$B_X$$ is convex if for all $$x,y \in B_X$$ and all $$\lambda \in [0,1]$$ we have

$$\lambda x + (1-\lambda)y \in B_X$$

From the definition of a unit ball we know

$$\|x\| \leq 1$$ and $$\|y\| \leq 1$$

Now for any $$\lambda$$ such that $$0 \leq \lambda \leq 1$$

$$|\lambda|\|x\|\leq 1$$ and $$|(1-\lambda)|\|y\|\leq 1$$

From the properties of the norm we can absorb the constants and define new upper bounds

$$\|\lambda x\| \leq |\lambda|$$ and $$\|(1-\lambda)y\|\leq |(1-\lambda)|$$

From this we use the Triangle Inequality to obtain

$$\|\lambda x + (1-\lambda)y\| \leq \|\lambda x\| + \|(1-\lambda)y\| \leq |\lambda| + |(1-\lambda)|$$

Which implies

$$\|\lambda x + (1-\lambda)y\| \leq 1 - \lambda + \lambda$$
$$\|\lambda x + (1-\lambda)y\| \leq 1$$

and so

$$\lambda x + (1-\lambda)y \in B_X$$

Please tell me this is right. If so can we start talking about Question 1?

16. Mar 13, 2005

### Data

Very good.

Alright, now #1. Can you tell me what the definition of $$B\left[x ; r\right]$$ is?

17. Mar 14, 2005

### Oxymoron

The definition of $$B[x;r]$$ is the closed ball centered on $$x \in X$$ of radius $$r$$.

And $$B_X$$ is the unit ball centered on $$x=0$$ with radius $$r = 1$$.

So $$rB_X$$ is just the unit ball centered at the origin but the radius is scaled by some number $$r$$.

Next, we make this a general ball by stating that the $$x$$ we choose need not be at the origin but at some other point.

Then the definition $$B[x;r] = x+rB_X$$ is just the scaled unit ball shifted to some arbitrary point.

So how do I begin to write this systematically?

18. Mar 14, 2005

### Oxymoron

Just while I'm here, I have another solved problem I'd like anyone to take a look at.

$$Question$$

If $$x$$ and $$y$$ are any two distinct norm-one vectors of an inner product space, show that

$$\left\|\frac{1}{2}(x+y)\right\| < 1$$

$$proof$$

Take any $$x,y \in V$$. Since $$x$$ and $$y$$ are norm-one vectors we have $$\|x\| = \|y\| = 1$$.

From the triangle inequality we have

$$\|x+y\| \leq \|x\| + \|y\|$$
$$\frac{1}{2}\|x+y\| \leq \frac{1}{2}\left(\|x\| + \|y\|\right)$$
$$\left\|\frac{1}{2}(x+y)\right\| \leq \frac{1}{2}(1+1)$$
$$\left\|\frac{1}{2}(x+y)\right\| \leq 1$$

Since $$x$$ and $$y$$ are distinct, ie. $$x \neq y$$, then

$$\left\|\frac{1}{2}(x+y)\right\| \neq 1$$

hence

$$\left\|\frac{1}{2}(x+y)\right\| < 1$$

19. Mar 14, 2005

### Oxymoron

The closed unit ball

$$B[x;r] = \{x\in X \, :\, \|x-x_0\| \leq r\}$$

That is, the norm of the vector from the origin to the point $$x$$ is at most $$r$$.

But for the unit ball $$B_X = \{x\in X\,:\, \|x\| \leq 1\}$$

we can say that

$$rB_X = |r|\|x_0\| \leq 1$$

which implies

$$rB_X = \|rx\| \leq |r|$$

So now we have an arbitrary radius of the ball centered at the origin, simply by multiplication by $$r$$. Now, the ball is still on the origin, to get it like the definition of an arbitrary closed ball, we need the origin to unspecific.

$$x+rB_X =$$

stumped.

Last edited: Mar 14, 2005
20. Mar 14, 2005

### Data

I think you meant

$$B\left[ x;r\right] = \{ z \in X : \| z - x \| \leq r \}$$

Can you now tell me exactly what #1 tells you to prove, in as simple terms as you can? This is usually a good step to starting a proof.

As to your completed question, it looks fine. There is one step that I would like to see justified better though. Why does $$x \neq y$$ imply $$|x+y| \neq |x| + |y|$$? If you don't know how to show this, here is a hint: Reread the statement of the Cauchy-Schwarz inequality.