Need Help with Inner Product Space X Questions?

[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 can't begin to prove them. Can anyone get me started? I only need pointers.

Thanks.

 

[Hurkyl]

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

 

[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 can't find anywhere a procedure on how to show that a subset is convex.??

 

[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~

 

[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?

 

[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!

 

[Data]

I will also note that

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

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).

 

[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.

doesn't make too much sense, unless we're talking about doesn't make too much sense, unless we're talking about \mathbb{R}.

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, I am going to stop now. I don't think I am making any sense. I think I am 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.

 

[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!

 

[Data]

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

 

[Oxymoron]

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

I still don't know where to go. I can't 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

 

[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.

 

[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

 

[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 :)

 

[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?

 

[Data]

Very good.

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

 

[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?

 

[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

 

[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.

 

[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.

 

[Data]

Since I probably won't be around for most of tomorrow, and undoubtedly you're asleep right now, I'll leave a couple more hints.

For the question you haven't done: if A and B are sets, then often, a good strategy to proving A=B is to first prove A \subseteq B and then to prove B \subseteq A.

For the question you've already done, when you're trying to justify the step I noted above, remember that if \circ is the inner product on X, then for a \in X, \ \mbox{we define} \ \|a\| \equiv \sqrt{ a\circ a}.

 

[Oxymoron]

Data, you are a legend. I have worked out the proof! I think. Let me just type it up.