# Thread: A few problems in functional analysis

1. Edit: I noticed the "sticky" article, so be at rest, these problems are not homeworks. When a problem is solved here I tell my prof that I got some "international help to solve that one" Hi there,

I'll post in this thread some problems I couldn't solve from the book: "Basic operator theory" by Goldberg and Gohberg. Any help/discussion is appreciated.

Here is one:

Chapter9 - problem 16

16. Prove that if a normed linear space X contains linearly
independent vectors x and y such that ||x+y|| = ||x|| + ||y||, then
there is a line segment contained in the unit sphere of X.
The converse is also true by the way (it's an exercise: problem 15), but I was able to prove it.

Thanks a lot!  2.

3. What about the line segment spanned by the unit vector in the direction of x + y?  4. Originally Posted by river_rat
What about the line segment spanned by the unit vector in the direction of x + y?
He is looking for a line segment in the unit sphere and not in just the unit ball. So he wants a segment consisting solely of points of norm 1. It is easy to get a line segment in the unit ball, since it is convex just connect any two points. It is more difficult to get them in the unit sphere -- for instance it is not possible in ordinary Euclidean space.  5. yes, that's the idea.

The thing is there are no other information about these vectors x and y.

In the special case where their norm is 1, that is, if they are on the unit sphere, I was able to prove that the set L={ax+(1-a)y} is indeed contained in the unit sphere for all dyadic fractions "a" in the interval [0,1]. And since this set is dense in [0,1], and since the unit sphere is closed, the whole line segment is in the unit sphere.

The thing is he doesn't say that the x and y are on the unit sphere. Of course, the natural thing to do would be to consider the vectors x1=x/||x|| and y1=y/||y|| but then things get ugly and I couldn't prove anything...  6. Originally Posted by Stranger
yes, that's the idea.

The thing is there are no other information about these vectors x and y.

In the special case where their norm is 1, that is, if they are on the unit sphere, I was able to prove that the set L={ax+(1-a)y} is indeed contained in the unit sphere for all dyadic fractions "a" in the interval [0,1]. And since this set is dense in [0,1], and since the unit sphere is closed, the whole line segment is in the unit sphere.

The thing is he doesn't say that the x and y are on the unit sphere. Of course, the natural thing to do would be to consider the vectors x1=x/||x|| and y1=y/||y|| but then things get ugly and I couldn't prove anything...
What you proved is about as far as I was able to get so far as well. You can always scale x and y simultaneously so that one or the other has norm 1. So, I think the trick may be to show that you can scale each one separately and preserve additivity of the norm -- but I don't see how to do that.

This will take more thought. It is more subtle than it appears at first sight. The converse is trivial -- if there is a line connecting two points x and y on the unit sphere then 1/2 x + 1/2 y is also of norm 1 and the result follows.  7. ah, i see now. where is that pencil and paper....  8. Originally Posted by Stranger
yes, that's the idea.

The thing is there are no other information about these vectors x and y.
One more embryonic thought. It may be instructive to think about two specific examples. Both are relate to 2 space. Consider the norms and max{ }

The first case yeilds the unit sphere as the unit diamond and the second as the unit square. Both contain line segments. The first is knows as the norm and the second as the norm  9. yes, I thought about these spaces too.

I also proved by the way that our space cannot be a Hilbert space. Also, it is impossible to have ||x-y|| = ||x|| - ||y||.

mmm I'm a bit confused. I was trying to write the proof in the case where they are on the unit sphere but got stuck. I admit it was my professor who thought that case would be solvable using dyadics... I'll try again.

But I thought about something: consider the sphere S = {u: ||u||=||x+y||}. If we could prove there is a line segment in this sphere (which should be, I think, exactly as difficult as proving the problem in the case ||x||=||y||=1) then this would imply that the unit sphere also contains a line segment, since the latter is merely a scalar multiple of the former.

What do you think?  10. Originally Posted by Stranger
yes, I thought about these spaces too.

I also proved by the way that our space cannot be a Hilbert space. Also, it is impossible to have ||x-y|| = ||x|| - ||y||.

mmm I'm a bit confused. I was trying to write the proof in the case where they are on the unit sphere but got stuck. I admit it was my professor who thought that case would be solvable using dyadics... I'll try again.

But I thought about something: consider the sphere S = {u: ||u||=||x+y||}. If we could prove there is a line segment in this sphere (which should be, I think, exactly as difficult as proving the problem in the case ||x||=||y||=1) then this would imply that the unit sphere also contains a line segment, since the latter is merely a scalar multiple of the former.

What do you think?
If x and y are unit vectors and then 1/2 x + 1/2 y is also a unit vector, i.e. the mid-point of the line joining x and y lies on the unit sphere. Now proceed inductively (the next step is to look at the segments from x to 1/2 x + 1/2y and from 1/2 x + 1/2 y to y) and you get the dyadic points along the line joining x and y to also lie on the unit sphere. Then you use the fact that the sphere is closed to conclude that the entire line is on the sphere.

There is a theorem for topological vector spaces that states that an open or closed set that is "mid-point convex" (contains the mid-points of segments joining points in the set) is in fact convex, and that is proved just as above using the dyadics.  11. You can do this special case just with the triangle inequality btw.

i. Now suppose then   i.e. If then consider where and proceed similarly.

Thus for all   12. Originally Posted by river_rat
You can do this special case just with the triangle inequality btw.

i. Now suppose then   i.e. If then consider where and proceed similarly.

Thus for all So now we have two proofs of the main theorem in the case in which the vectors x and y are linearly independent and of norm 1.

From that special case we can prove the main result. It will follow if we can show that given x and y linearly indepedent and such that we can find such vectors of norm 1. We know that if x and y satisfy our condition then for an a>0 so do ax and ay. So now suppose that we have such vectors x and y. Assume that then by division by we may assume that and The general theorem will now follow if we can show that for . We now prove this.

Let and suppose x and y satisfy . Then and   Hence and therefore which completes the proof.  13. Thanks very much both of you!

I'm sorry I couldn't reply earlier; my computer keeps crashing down. I guess it's time for me to buy a laptop. But these aren't perfect either... whatever, if I don't reply immediately, it only means I'm unable to reply because of that dumb pc.

These problems are not homeworks by the way. It's just that I feel bad when I've tried hard to find a solution and then, not only reach nothing, but also never know the solution...

ok, here is another one. I've hesitated for a long time to post that one because I really feel it's an easy one, but I keep trying and don't reach the thing.

It's problem 22 the same chapter.

22. Let M be a closed subspace of X. Define the quotient space
X/M to be the vector space consisting of the cosets [x] =
{x+z : z  M} with vector addition and scalar multiplication
defined by [x] + [y] = [x+y], a[x] = [ax]. Define ||[x]||_1 =
d(x,M), the distance from x to M. Prove that ||.||_1 is a
norm on X/M and that (X/M,||.||_1 ) is complete.
(d(x,M) is of course the infimum of all d(x,y) for y in M)

I've proved the norm, but I'm a bit stuck with completeness. As usual I assumed {[x]n}n was a Cauchy sequence. Then I finally reached that given eps>0,
d(xn-xm,M)<eps for all n,m>n0. _____(*)
where [xn]=[x]n

Then I claimed that this implies (xn-xm) is in the closure of M (in which case the proof would be finished: M is closed, thus (xn-xm) is in M, thus [xn-xm]=0, thus [xn]=[xm] for all n,m>n0, thus [x]n is eventually stationary, thus convergent)

It turns out I was wrong however, my prof says (*) does not imply that (xn-xm) is in the closure.

Any little hint?

Oh, and another thing. In a problem I reached the conclusion that 2 vectors x,y must satisfy:
Re <x,y> = 0
where <x,y> is their inner product. Can I translate this condition into something else on x and y themselves, instead of their product? Or is it all I can say?

Thanks a lot  14. Originally Posted by Stranger
Thanks very much both of you!

I'm sorry I couldn't reply earlier; my computer keeps crashing down. I guess it's time for me to buy a laptop. But these aren't perfect either... whatever, if I don't reply immediately, it only means I'm unable to reply because of that dumb pc.

These problems are not homeworks by the way. It's just that I feel bad when I've tried hard to find a solution and then, not only reach nothing, but also never know the solution...

ok, here is another one. I've hesitated for a long time to post that one because I really feel it's an easy one, but I keep trying and don't reach the thing.

It's problem 22 the same chapter.

22. Let M be a closed subspace of X. Define the quotient space
X/M to be the vector space consisting of the cosets [x] =
{x+z : z  M} with vector addition and scalar multiplication
defined by [x] + [y] = [x+y], a[x] = [ax]. Define ||[x]||_1 =
d(x,M), the distance from x to M. Prove that ||.||_1 is a
norm on X/M and that (X/M,||.||_1 ) is complete.

(d(x,M) is of course the infimum of all d(x,y) for y in M)

I've proved the norm, but I'm a bit stuck with completeness. As usual I assumed {[x]n}n was a Cauchy sequence. Then I finally reached that given eps>0,
d(xn-xm,M)<eps for all n,m>n0. _____(*)
where [xn]=[x]n

Then I claimed that this implies (xn-xm) is in the closure of M (in which case the proof would be finished: M is closed, thus (xn-xm) is in M, thus [xn-xm]=0, thus [xn]=[xm] for all n,m>n0, thus [x]n is eventually stationary, thus convergent)

It turns out I was wrong however, my prof says (*) does not imply that (xn-xm) is in the closure.

Any little hint?

Oh, and another thing. In a problem I reached the conclusion that 2 vectors x,y must satisfy:
Re <x,y> = 0
where <x,y> is their inner product. Can I translate this condition into something else on x and y themselves, instead of their product? Or is it all I can say?

Thanks a lot
The proof is not hard, but there are a few steps involved. It is a standatd result, unlike your earlier question. But I am a bit slow with Tex, so rather than try to type up the proof I will give you a reference. Take a look at the book Functional Analysis by Kosaku Yosida. In particular look at Chapter 1 section 11, Factor Spaces of a B-Space (my copy happens to be the Fourth Edition). Yosida is a pretty standard reference in general Functinal Analysis and you would do well to get a copy if you are interested in the area.

You can also find a proof in Walter Rudin's Functional Analysis which is, in my opinion the best text on the subject to be found. That is the text from which I originally learned the subject. But beware, because Rudin's proofs tend to be very slick, so you need to study them a bit to get a feel for the intuition behind them.  15. mmm that's a bit confusing: in both books you mentioned, the space X is assumed to be a Banach space, that is complete. My stupid book forgot to mention that (and I was indeed in need of the completeness of X). Well, my book isn't bad after all, but the author often forget to mention a few givens, like here the completeness... I'll try again then.

Ok, here is a nice one:

20. Given two spheres ||x-y0|| = ||y0|| and ||x+y0|| = ||y0|| in a normed
linear space, how many points can these spheres have in common?
I proved a few things here:
1) They'll always share the point 0.
2) They may share an uncountable number of points (for instance, with the linf norm).
3) They can never be identical unless y0=0, in which case they both coincide with the zero vector.

But now, could they share for instance 2 points?  16. Originally Posted by Stranger
mmm that's a bit confusing: in both books you mentioned, the space X is assumed to be a Banach space, that is complete. My stupid book forgot to mention that (and I was indeed in need of the completeness of X). Well, my book isn't bad after all, but the author often forget to mention a few givens, like here the completeness... I'll try again then.
Since you were to prove the completion of the quotient space you must have completeness of X. For if you consider the trivial case in which the subspace is the 0 subspace then the quotient is isomorphic to X.

Since one rarely works with normed spaces that are not complete, the omission is understandable.  17. whatever...

I've also proved that the balls can only share points on their boundaries, that is, if x is an interior point of one ball, it cannot be on the other (closed) ball. Though this fact is about balls not spheres, I can't help thinking this implies they cannot intersect in a countable number of points (except just 1 point of course). That's because of the geometric interpretation...

But it still has to be proved, I'll try to write one.

By geomerty, I'm sure I'll have to make use of the convexity of the balls. I think I'm close...  18. Originally Posted by Stranger
whatever...

I've also proved that the balls can only share points on their boundaries, that is, if x is an interior point of one ball, it cannot be on the other (closed) ball. Though this fact is about balls not spheres, I can't help thinking this implies they cannot intersect in a countable number of points (except just 1 point of course). That's because of the geometric interpretation...

But it still has to be proved, I'll try to write one.

By geomerty, I'm sure I'll have to make use of the convexity of the balls. I think I'm close...
If you have shown that the intersection of the balls is the intersection of the spheres, then it follows that the intersection of the spheres is itself convex. Hence if it contains two points it also contains the line segment joining them and cannot contain only a countable number of points, unless the intersection is a single point.  19. Sorry again (dumb pc).

I completed the problem about an hour after my last post.

Well that's all for now thanks very much.

I'll try hard before posting the next one   20. Hi there, long time no see .

I added a comment on my first post as I noticed the "sticky" article about homeworks.

Well, things went well during that time. I solved 55 problems from chapter one, which is about Hilbert spaces . Well, I actually solved 53, 2 were solved by my prof.

Here is one we couldn't solve however: I will denote the closure of the span by a bold sp

56. Let {u1,u2,...} be a set of vectors and let {v1,v2,...} be an orthogonal system. We call {v1,v2,...} a backward orthogonalization of {u1,u2,...} if sp{uj,uj+1,... } = sp{vj,vj+1,...}for j = 1,2,... .
Prove that there exists a backward orthogonalization for {u1,u2,...} if and only if for every j we have uj not in sp{uj+1,uj+2,...}
We could prove the forward direction, that is, if we have a back ortho, then for every j, uj not in sp{uj+1,uj+2,...} . But what about the reverse? How can we create a back ortho, if for every j we have uj not in sp{uj+1,uj+2,...} ?

Thanks a lot  21. I've just solved the problem if anyone is interested.

The trick is to do exactly the inverse of the Gram-Schmidt procedure:

Let Mj=sp{uj+1,uj+2,...} and let

vj = uj - PMj(uj) where PMj(uj) is the projection of uj on Mj. Then you can prove the vj are orthogonal and span the thing. The condition uj not in Mj turned out to be for the spanning part actually, not the orthogonalization part.

I'm saying "exactly the inverse of the Gram-Schmidt procedure" , but not quite... in the Gram-Schmidt, you subtract its projection on the new vectors, not the old ones... this would be obviously impossible here.

Thanks for those who thought about it   22. Hi again,

I've been trying to find the intersection of any 2 spheres in a normed space, but I failed. In an inner product space, I was able to get it:
Let S1={v:||v-x||=R1} and S2={v:||v-y||=R2} and assume ||x-y||=L.
Then S1^S2={v:||v-(R2x+R1y)/(R1+R2)||^2=R1R2(1-(L^2)/((R1+R2)^2)) and
2Re<v,x-y>=R2^2-R1^2+||x||^2-||y||^2}

Sorry if I'm using old fashioned symbols, I'm just not used to TeX. If anything is not clear, please mention it. As expected, in the special case where R1=R2 and ||x||=||y|| and the space is real, the intersection is a sphere is the space orthogonal to x-y.

Anyway, so I'm unable to get the intersection in normed spaces in general. I don't know, my failed attempts make me think the thing will be really tricky... but geometrically speaking, the intersection is not so bad if you consider the sup norm or the 1-norm for instance... it shouldn't look bad...

If anyone can help it will be great :)

If that intersection is too ambitious, I want at least to prove that, in case ||x-y||=L<R1+R2, then the intersection is nonempty. If L=R2+R2, the matter is simple, take v=(R2x+R1y)/(R1+R2). But if L<R1+R2 I don't know....

By the way, even though I got the above intersection in an inner product space, I still couldn't get a point which satifsfies both conditions, that is, I couldn't prove the above intersection is non-empty when L<R1+R2 (so, same problem there).

Thanks very much!  Bookmarks
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