1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cauchy sequences in an inner product space

  1. Mar 19, 2005 #1
    Im in need of some guidance. No answers, just guidance. :smile:


    Let [tex](x_m)[/tex] be a Cauchy sequence in an inner product space, show that


    is bounded.


    From the definition we know that all convergent sequences are Cauchy sequences. But it is not true that the converse always holds. However, in normed linear spaces the converse is always true. That is our Cauchy sequence is convergent to a point [tex]x[/tex].

    Since [tex](x_m)[/tex] is a convergent sequence we take an [tex]\epsilon[/tex] such that

    [tex]\|x_m - x\| \leq \epsilon[/tex] if [tex]m > N[/tex].

    Now we fix [tex]N[/tex] so that

    [tex]\|x_m\| \leq \|x_m - x\| + \|x\| < \epsilon + \|x\|[/tex]

    which holds for all [tex]m > N[/tex]. This is just a matter of adding [tex]\|x\|[/tex]

    Now we define

    [tex]M = \mbox{max}\{\|x_1\|,\|x_2\|,\dots,\|x_N\|, (1+\|x\|)\}[/tex]

    Then [tex]\|x_m\| < M \, \forall \, m[/tex]

    That is, the sequence is bounded.
  2. jcsd
  3. Mar 19, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure? What about, say, Q, which forms a normed linear space over itself, and we all know that Q is full of Cauchy sequences that don't converge...

    The first approach I can see is to review the proof that Cauchy sequences converge in R and adapt it to your purposes.
  4. Mar 19, 2005 #3
    You probably know what my first request is going to be already :P

    What's the definition of a Cauchy sequence (and don't just give me a limit, because that has its own definition too! State it in terms of the definition of a limit)?

    With the help of the "other" triangle inequality, [tex] \| y - z \| \geq \|y\| - \|z\|[/tex], see if you can get anywhere from that definition.
  5. Mar 20, 2005 #4
    You guys, Data and Hurkyl, do you just look at my questions and instantly know how to do them? I really appreciate your wisdom and effort.

    So I guess my method of solving it is flawed because I used assumed the Cauchy sequence converges to some point then used the fact that it converges to show that it must be bounded above by some number K.

    One question before I go on. In regards to this question, does [tex]\left\{\|x_n\| : n = 1, \dots, \infty\right\}[/tex] form a subset of an inner product space?

    Ok, Data, starting from basics.


    [tex](x_n)[/tex] is said to converge to a point [tex]x_0 \in X[/tex] (has a limit [tex]x_0[/tex]) if for all [tex]\epsilon > 0 \, \exists \, n_0 \in \mathbb{N}[/tex] such that

    [tex]\|x_n - x_0\| < \epsilon[/tex]

    or equivalently

    [tex]x_n \in B(x_0;\epsilon)[/tex]

    We want to show that
    [tex]\left\{\|x_n\| : n = 1, \dots, \infty\right\}[/tex]
    is bounded. If we can show that [tex](x_m)[/tex] is convergent then it is bounded. Am I correct in assuming that the sequence [tex](x_m)[/tex] is formed by taking consecutive [tex]\|x_n\|[/tex] for [tex]n = 0, 1, \dots, \infty[/tex]?

    Let [tex]x_0 = \lim x_n[/tex], then [tex]\exists \, n_0 \in \mathbb{N}[/tex] such that [tex]\|x_n - x_0\| < \epsilon \, \forall \, n \geq n_0[/tex]

    The picture you should have in your mind is a closed ball centered on [tex]x_0[/tex] (the limit) with radius [tex]\epsilon[/tex]. As you follow the sequence eventually all the [tex]x[/tex]'s will stay inside the ball of radius [tex]\epsilon[/tex]

    So now we take the radius of the ball

    [tex]r = \max\{\epsilon, \|x_1\|, \|x_2\|, \dots, \|x_{n-1}\|\}[/tex]


    [tex]\|x_n - x_0 \| \leq r[/tex]

    so it is bounded by [tex]r[/tex]
  6. Mar 20, 2005 #5
    I don't instantly know how to do them, of course! If I'm lucky sometimes I can figure it out after a few minutes of looking at it though (for example, I had to look up what a Cauchy sequence was, although I did guess based on the Cauchy criterion for convergence of real sequences. I've never used them before!) :).

    As to your first question, [tex]\{ \|x_n\| : n = 1, 2, ... \}[/tex] is a subset of an inner product space. Namely, of the real numbers. Remember that [tex]\| x \|[/tex] is just a real number! On the other hand, [tex] \{ x_n : n = 1, 2, ... \}[/tex] is a subset of the inner product space mentioned in the statement of your problem.

    Anyways, your definition is a little off. Try this:

    If [itex](x_n)[/itex] is a sequence with elements in an inner product space, then we say that

    [tex]( \ (x_n) \ \mbox{is a Cauchy sequence} \ ) \Longleftrightarrow (\ \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ n, \ m \geq N \Longrightarrow \|x_n - x_m\| < \epsilon \ )[/tex]

    Your assumption is almost correct. According to your question, there is one [itex]x_n[/itex] for each [tex] n=1, 2, 3, ...[/tex].

    Now, according to the definition I posted above, we can just choose [tex] m = N[/tex] for a particular [itex]\epsilon[/itex] (we can choose the [itex]\epsilon[/itex]). By doing this, what can I say about [tex]\| x_n \|[/tex] for every [tex]n > m[/tex]? Remember, all you can use are definitions, and known results (like the inequality I mentioned in my first post, which I think you will find in handy).
    Last edited: Mar 20, 2005
  7. Mar 20, 2005 #6
    Just a little clarification: From your last post, you seem to be a little bit confused about the difference between the notion of convergence and a Cauchy sequence. Unfortunately, for this problem, we can't prove that [tex](x_n)[/tex] is convergent (because it isn't necessarily).
    Last edited: Mar 20, 2005
  8. Mar 20, 2005 #7
    [tex]\|x_n\| \leq \|x_n - x_m\| + \|x_m\|[/tex]
  9. Mar 20, 2005 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    By the way, R is an inner product space, and the sequence of the ||x_i|| lives in R. (Can you show the ||x_i|| form a Cauchy sequence in R? If so, you know R is complete, and thus this sequence converges, even if the {x_i} don't)
  10. Mar 20, 2005 #9
    Correct. And what can you replace the [tex]\|x_n - x_m\|[/tex] by while keeping the inequality valid (something that you get to choose...)?

    Alternatively, Hurkyl's suggestion also works out very nicely~
  11. Mar 20, 2005 #10


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The more familiar you become with things, the easier they become.

    For instance, I know how Cauchy sequences "behave" -- they don't head off towards infinity, and they can't have multiple limit points... a Cauchy sequence that doesn't converge behaves as if it's converging to something "missing".

    More precisely, there's a bigger metric space (called the Cauchy completion) that contains your space, but every Cauchy sequence converges.

    (As an example, R is the Cauchy completion of Q)

    So, even if your Cauchy sequence doesn't converge, I know it should behave like it does, and in particular you should be able to find a bound for the sequence.

    The easiest way I could imagine for finding that bound would be to appropriately modify the proof that Cauchy sequences converge in complete spaces like R^n -- one of the steps involved is bounding tails of the sequence.
  12. Mar 20, 2005 #11
    [tex]\|x_n\| \leq \epsilon + \|x_m\|[/tex]
  13. Mar 20, 2005 #12
    Right, and like I said, you can just choose [tex] m = N[/tex]. Since [tex]\| x_N \| [/tex] is just a real number, you should be missing just a couple of minor details to finish the proof. I'm sure you can figure those out~
    Last edited: Mar 20, 2005
  14. Mar 20, 2005 #13
    [tex]\|x_n\| \leq \epsilon + \|x_M\|[/tex]
    [tex]\|x_n\| - \|x_M\| \leq \epsilon[/tex]
    [tex]\|x_n - x_M\| \leq \epsilon[/tex]

    So by the definition the Cauchy sequence converges to [tex]x_M[/tex].
  15. Mar 20, 2005 #14
    Not quite. Remeber, we chose a particular epsilon, which is not enough to say that anything converges. The important point here is that [itex] \| x_n \| \leq \epsilon + \| x_N \| [/itex] holds [itex] \forall n \geq N[/itex], ie. you've bounded [itex] \| x_n \|[/tex] for [itex] n \geq N[/itex]. The details I referred to simply have to do with the terms [itex] x_n [/itex] with [itex] n < N[/itex].
  16. Mar 20, 2005 #15
    So have we proved that [tex]\|x_n\|[/tex] is bounded?
  17. Mar 20, 2005 #16
    Here's what we have so far:

    From the definition of a Cauchy sequence, we know that

    [tex]\lim_{\mbox{min}(m, n) \rightarrow \infty} \| x_n - x_m\| = 0[/tex]

    or in other words

    [tex] \exists N \forall \epsilon >0 \ \mbox{s.t.} \ n, m \geq N \Longrightarrow \|x_n - x_m\| < \epsilon[/tex]

    Thus, if we choose a particular [itex] \epsilon >0 [/itex], we can find a particular N such that

    [tex] \forall n, m \geq N, \ \|x_n - x_m \| < \epsilon[/tex]

    and thus

    [tex] \forall n \geq N, \ \|x_n - x_N \| < \epsilon[/tex]

    and since, by the inequality I mentioned above, [itex]\forall n \geq N[/itex],

    [tex]\|x_n\| - \|x_N\| \leq \| x_n - x_N \| < \epsilon \Longrightarrow \|x_n\| < \epsilon + \|x_N\| = M \in \mathbb{R}[/tex]

    so we have bounded [itex] \|x_n\| \forall n > N[/tex]. I'm sure you can fill in the rest (which is trivial!) :smile:
  18. Mar 20, 2005 #17
    On the other hand, here's a possible argument using Hurkyl's idea:

    As in my last post, we know that

    [tex]\exists N \ \forall \epsilon>0 \ \mbox{s.t.} \ n, m \geq N \Longrightarrow \|x_n - x_m \| < \epsilon[/tex]

    but by the inequality I mentioned,

    [tex]\|x_n\| - \|x_m\| \leq \|x_n - x_m\|[/tex]

    and similarly, since [itex] \|x_n - x_m\| = \|x_m - x_n\|[/itex],

    [tex] -(\|x_n\| - \|x_m\|) = \|x_m\| - \|x_n\| \leq \|x_n - x_m\|[/tex]

    so we can certainly conclude that

    [tex]| \|x_n\| - \|x_m\| | \leq \| x_n - x_m \|[/tex]

    which gives us

    [tex] \exists N \ \forall \epsilon >0, \ \mbox{s.t.} \ n, m \geq N \Longrightarrow | \|x_n\| - \|x_m\| | < \epsilon[/tex]

    and thus [itex] (\|x_n\|)[/itex] is also a Cauchy sequence, over the real numbers. But since Cauchy sequences over [itex]\mathbb{R}[/itex] always converge, we find that [itex](\|x_n\|)[/itex] converges and is thus bounded.
  19. Mar 21, 2005 #18
    I hope I dont disappoint you too much but to me it looks like [tex]\|x_n\|[/tex] is bounded.
  20. Mar 21, 2005 #19
    it certainly is! Not getting a problem isn't any reason to be dissapointed... everyone has trouble with some problems! :smile:
  21. Mar 21, 2005 #20
    I have another question I'd like your help on.


    Let [tex](x_n)[/tex] be a sequence of elements on a Hilbert space [tex]H[/tex] for which


    Show that the sequence of partial sums

    [tex]s_n = \Sigma_{k=1}^{n}x_n[/tex]

    converges to a point in [tex]H[/tex].


    Since we know that [tex]H[/tex] is a Hilbert space, we know that every Cauchy sequence in [tex]H[/tex] is convergent. So to answer this question we have to prove that [tex](x_n)[/tex] is Cauchy.

    [tex]\|x_n - x_m\|^2 \rightarrow x[/tex] as [tex]n,m \rightarrow \infty[/tex]

    [tex]\|x_n - x_m\|^2 = \Sigma_{j=1}^{\infty}|x_{nj} - x_{mj}|^2 \geq |x_{nj} - x_{mj}|^2[/tex]


    [tex]|x_{nj} - x_{mj}| \rightarrow x[/tex] as [tex]j \rightarrow \infty[/tex]

    Therefore [tex](x_{nj})[/tex] is a Cauchy sequence convergent to [tex]x[/tex]

    But is [tex]x \in H[/tex]? Now we need to show that [tex]x \in H[/tex] and [tex]x_{n} \rightarrow x[/tex]

    The partial sum [tex]\Sigma_{j=1}^{n}|x_j|^2 = \Sigma_{j=1}^n|\lim_{k\rightarrow\infty}x_{j}|^2[/tex]
    [tex]= \lim_{k\rightarrow\infty}\Sigma_{j=1}^{n}|x_{nj}|^2[/tex]


    [tex]\Sigma_{j=1}^m|x_{nj}|^2 = \|x_n\|^2 \in H[/tex]


    [tex]\lim_{k\rightarrow\infty}\left(\Sigma_{j=1}^{\infty}|x_{nj}|^2\right)= \lim_{k\rightarrow\infty}\|x_n\|^2 \leq \lim_{k\rightarrow\infty}M[/tex]

    where [tex]\|x_n\|^2 \leq M \, \forall \, k[/tex]

    We know that such an [tex]M[/tex] exists because Cauchy sequences are bounded. So the partial sum

    [tex]\Sigma_{j=1}^{\infty}|x_n|^2 = \lim_{k\rightarrow\infty}\|x_n\|^2 \leq M[/tex]

    This is an increasing sequence that is bounded above. So

    [tex]\Sigma_{j=1}^{\infty}|x_n|^2 < \infty[/tex]

    ie [tex]x \in H[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Cauchy sequences in an inner product space
  1. Inner Product Spaces (Replies: 2)

  2. Cauchy sequence (Replies: 3)

  3. Cauchy sequence (Replies: 3)