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Continuity in Normed Vector Spaces

  1. Oct 26, 2008 #1
    The problem statement, all variables and given/known data
    Let V and V' be real normed vector spaces and let f be a linear transformation from V to V'. Prove that f is continuous if V is finite dimensional.

    The attempt at a solution
    Let [tex]v_1, v_2, \ldots, v_n[/tex] be a basis for V, let e > 0 and let v in V. I must find a d such that for all u in V such that ||u - v|| < d, ||f(u) - f(v)|| < e.

    Now [tex]f(u) - f(v) = a_1 f(v_1) + \cdots + a_n f(v_n)[/tex] for some scalars [tex]a_i[/tex]. Thus [tex]\|f(u) - f(v)\| \le \|a_1 f(v_1)\| + \cdots + \|a_n f(v_n)\|[/tex].

    If [tex]\|a_i f(v_i)\| < e/n[/tex], then ||f(u) - f(v)|| < e. This is as far as I've gone. Any tips?
     
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  3. Oct 26, 2008 #2

    morphism

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    Hint: give V and V' the Euclidean norm. Why can we do this without loss of generality?
     
  4. Oct 27, 2008 #3
    We can do this because all norms on a finite-dimensional vector space are equivalent. I know this is because the next problem after this one asks me to prove this fact. I feel it would be cheating if we proceed in this manner.

    In any case, I don't understand how the Euclidean norm will help. The main issue, as I'm perceiving it, is that I don't know anything about f apart from the fact that it is linear. Hence, I'm unable to figure out how ||u - v|| < d would imply ||f(u) - f(v)|| < e, even if I knew the right d. I guess that I will need some kind of bound on ||f(v)|| in terms of ||v||.
     
  5. Oct 27, 2008 #4

    Dick

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    Start by proving continuity at the origin. ||w||<d -> ||f(w)||<e. The easy way to do this is to observe that the unit sphere ||w||=1 is compact in a FINITE DIMENSIONAL space. What does that tell you about ||f(w)|| on the sphere?
     
  6. Oct 27, 2008 #5
    Good idea. And if we show continuity at the origin, continuity everywhere follows.

    This I've been trying to prove on a related question. All the proofs I've seen of this rely on the fact that a finite dimensional space is isomorphic to R^n. Is there a proof that doesn't use this fact?

    I presume you want me to say that it is bounded. I know this is the case if f is continuous, but f is not continuous.
     
  7. Oct 27, 2008 #6

    Dick

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    Yes, I did want you to say bounded. But you are right. That needs f to be continuous.
     
  8. Oct 27, 2008 #7
    I'm getting the impression that there is no way of proving that {w : ||w|| = 1} is compact without showing that it is isomorphic to some a subset of R^n and then invoking the Heine-Borel theorem, unless of course I redo the Heine-Borel theorem for finite-dimensional normed vector spaces.
     
  9. Oct 27, 2008 #8

    Dick

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    Don't worry about that. That was not a very good suggestion. Though it's certainly true that {||x||=1} is compact. The real point here is that if you pick a basis for V, {v1,...,vn} and if v=a1*v1+...+an*vn and ||v||<d, that you can find some kind of bound on |a1|,|a2|,...,|an|. That's what good having the freedom to move to the Euclidean norm would do you. It would make that step easy. Sorry, I haven't had more time to think about this.
     
    Last edited: Oct 27, 2008
  10. Oct 27, 2008 #9

    Dick

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    Ok, think about two dimensions. Pick two sequences a1n and a2n, v=a1n*v1+a2n*v2 and ||v||<d. Suppose |a1n| is unbounded going to infinity. That gives you |a1n|*||v1+a2n*v2/a1n||<d. Use that ||a+b||>=|||a||-||b|||. If a2n/a1n were also unbounded that contradicts the limiting upper value of d. Hence a2n/a1n is bounded. Pick a subsequence converging to say, c. If c=0, that's another contradiction. If c is not equal to zero then v1 and v2 are linearly dependent, right? Hence a1n and a2n are bounded and if ||a1*v1+a2*v2||<d then a1 and b2 are bounded. I think this whole thing is really similar to the proof all finite dimensional norms are equivalent. It's a pity those two proofs weren't in the opposite order.
     
  11. Oct 28, 2008 #10
    From |a1n| * ||v1 + a2n / a1n * v2|| < d we get ||v1 + a2n / a1n * v2|| < d / |a1n| and using ||a + b|| >= |||a|| - ||b||| on the left-side of the last inequality yields

    ||v1|| - |a2n / a1n| * ||v2|| < d / |a1n| and |a2n / a1n| * ||v2|| - ||v1|| < d / |a1n|.

    Taking limits, the right side of each inequality becomes 0, hence

    ||v1|| - ||v2|| * lim |a2n / a1n| < 0 and ||v2|| * lim |a2n / a1n| - ||v1|| < 0

    so

    ||v1|| / ||v2|| < lim |a2n / a1n| and lim |a2n / a1n| < ||v1|| / ||v2||.

    But this is impossible, so lim |a1n| is not infinity. We can use the same argument on |a2n| to show that it doesn't diverge.

    OK. Where exactly are we going with this?

    What proof are you referring to? The proof I have uses the compactness of the unit sphere and the fact that a function on a compact set has a minimum and maximum to show that two norms are equivalent on the unit sphere. From there, it is easy to show that two norms are equivalent everywhere.
     
  12. Oct 28, 2008 #11

    Dick

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    Where we are going is:||a1*v1+a2*v2||<d and we now know there is an M(d) such that |a1|<=M and |b1|<=M. It should be pretty easy to write down a bound for ||f(a1*v1+a2*v2)||. I wasn't referring to any special proof of norm equivalence - I was just guessing it would have a part that looked like this. I'm a little dubious about your proof of boundedness. Where do you use that v1 and v2 are linearly independent? You certainly have to use that. lim x->inf ||x*v1-x*v1||=0. There may be an easier way to go about this whole thing. But that's all I could think of. Maybe morphism knows a better trick.
     
  13. Oct 28, 2008 #12
    Ah, right.

    I didn't really prove that |an1| and |an2| are bounded. I only showed that there limit is not infinity since that was our initial assumption. In any case, I like this approach and will try to make it work to my satisfaction.
     
  14. Oct 28, 2008 #13
    On second thought, if their limits are not infinity, then they must be bounded.

    Also, my proof of boundedness is unnecessarily long: Upon taking the limit of ||v1 + a2n / a1n * v2|| < d / |a1n|, we get

    lim ||v1 + a2n / a1n * v2|| < 0

    which is impossible. In any case, it is all wrong. The error is that strict inequalities are not preserved by limits. For example, 0 < 1/n for all positive integers n. However lim 0 = 0 < lim 1/n = 0 does not hold.
     
    Last edited: Oct 28, 2008
  15. Oct 28, 2008 #14

    Dick

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    A priori, the limit of a2n/a1n may not exist (you'd need to use the assumption v1 and v2 are linearly independent). But if you can show a2n/a1n is bounded (and you surely can) you can extract a convergent subsequence -> c. And then, yes, ||v1+c*v2||<=0. So?
     
  16. Oct 28, 2008 #15
    Oh right. I don't understand how linear independence plays a role.

    I understand where the c comes from but where did this inequality come from?
     
  17. Oct 28, 2008 #16

    Dick

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    YOU said, ||v1+(a2n/a1n)v2||<d/|a1n|. |a1n|->infinity. You agree you can pick a subsequence of a2n/a1n that converges to some number c. Where do YOU think the inequality comes from? The whole point here is that if v1 and v2 ARE linearly dependent, you CAN have ||a1n*v1+a2n*v2||<d AND |an1|->infinity and |an2|->infinity. That's what I'm are trying to avoid. You somehow have to use linear independence.
     
  18. Oct 28, 2008 #17
    Oh yeah. Sorry about that. My thinking is slower than usual today.
     
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