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Let $\mathbb{P}$ denote the space of all finite degree polynomials in one variable. Show that $\mathbb{P}$ is never complete with respect to $P_1$ norm, i.e., $\|\cdot\|_1$ by giving an example Cauchy sequence that does not converge inside $\mathbb{P}$?

Saeed
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    One important note : With respect to what norm ? – Rebellos Feb 19 '19 at 20:28
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    wrt any norm in vector space – Saeed Feb 19 '19 at 20:30
  • I think that one counterexample which may interest you is the sequence of polynomials which defines the Taylor series for $e^x$ – b00n heT Feb 19 '19 at 20:33
  • I corrected the statement. – Saeed Feb 19 '19 at 20:34
  • Thank you b00n heT. – Saeed Feb 19 '19 at 20:39
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    @Saeed: Since this is an infinite-dimensional vector space, there are lots of non-equivalent norms, that means the truth of the statement can differ from norm to norm. Do you know (or suspect) that your statement is true for any norm? – Ingix Feb 19 '19 at 20:45
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    Please put your question in the body of your question and not just in the title. While you are doing that, please clarify the question to address the comments discussed above and say something about your work so far. – Rob Arthan Feb 19 '19 at 20:48
  • @Ingix It turns out that the set of polynomials is never complete, regardless of the norm we define on it. It follows from the triangle inequality and the fact that the polynomial $x$ is a non-zero element. – stressed out Feb 19 '19 at 20:56
  • @stressedout: you are making assumptions about the kinds of metric that are allowed (the question doesn't say the metric arises from a norm on the underlying vector spaces). The OP needs to clarify this question. – Rob Arthan Feb 19 '19 at 21:00
  • @RobArthan The OP mentions in the comments that he is considering "norms". See the fourth comment please. Plus, Ingix was talking about norms too. Hence, my comment that the statement is true regardless of the norm defined on the space. – stressed out Feb 19 '19 at 21:01
  • @stressedout: the OP should edit the question and not waste our time by making us work through all the comments to find out what is actually being asked. As there is no evidence of any work on the OP's part, I am voting to close. – Rob Arthan Feb 19 '19 at 21:05
  • @RobArthan Yes, I agree with you that the OP should edit the question. The OP did edit the question once, but it wasn't a good edit really. I think they should edit the question again and make the question self-contained. – stressed out Feb 19 '19 at 21:06
  • https://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-hamel-basis-of – Eric Wofsey Feb 19 '19 at 23:03
  • @XanderHenderson Nice to meet you again, Xander Handerson. :P Thank you for the counter-exmaple. Does trivial metric come from a norm? What does the norm look like? $|x|=1$ for any non-zero $x$ or something like that? This one seems to fail homogeneity. – stressed out Feb 19 '19 at 23:21

1 Answers1

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A correct solution:

The set $\{1,x,x^2,\cdots\}$ is a basis for $\mathbb{P}$. Consider the subspaces $X_n= \langle 1, x, \cdots, x^n \rangle$. Each $X_n$ is closed because every finite dimensional normed space is complete. Also, $X_n$ is a proper subset of $X$ and has empty interior because if it contains an open ball, the homogeneity of the norm allows it to cover all elements in $X$. Also, $X = \cup_{n=1}^{\infty} X_n$. Therefore, $X$ cannot be complete because otherwise we have just shown that it is a countable union of nowhere dense sets, which contradicts Baire's category theorem.

The initial answer which is wrong (even though I loved it):

Given any norm $\|\cdot\|$ on $\mathbb{P}$, the statement is true. Since it is a norm, and the polynomial $B_k(x)=x^k$ is a non-zero polynomial for all $k \in \{0,1,\cdots\}$, the real number $\|x^k\|$ is not zero for any $k$. Now consider the sequence $$P_n(x) = \sum_{k=0}^n \frac{1}{\|x^k\|2^k}x^k$$

By the triangle inequality, we have that

$$\|P_n(x)-P_m(x)\| = \left\|\sum_{k=m}^n \frac{1}{\|x^k\|2^k}x^k \right\|\leq \sum_{k=m}^n\frac{1}{2^k\|x^k\|}\|x^k\|=\sum_{k=m}^n\frac{1}{2^k}$$

The last sum can be made arbitrarily small because $\sum_{k=0}^{\infty}\frac{1}{2^k}=2$. Hence, the sequence $\big(P_n(x)\big)_{n=1}^{\infty}$ is Cauchy but it turns out, as pointed out by Eric Wofsey, that whether it converges in $\mathbb{P}$ or not is not a simple matter.

stressed out
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    +1: This is a good answer to a badly asked question. More generally, this argument shows that the union of an increasing family of finite-dimensional subspaces of a normed space cannot be complete. The dimension (qua vector space) of a Banach space cannot be countably infinite. – Rob Arthan Feb 19 '19 at 21:09
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    Nice solution, and the fix was 'simple'! Upvoted, and I'll delete my comments, as they are irrelevant now. – Ingix Feb 19 '19 at 22:09
  • @Ingix Thank you. Your comment was crucial in improving the answer. I appreciate it. – stressed out Feb 19 '19 at 22:11
  • This is wrong: you have never proved that the sequence does not converge, and in fact it does converge for some norms. – Eric Wofsey Feb 19 '19 at 22:44
  • @EricWofsey Do you mean that the speed of $|x^k|$ becoming small might make it converge? – stressed out Feb 19 '19 at 22:45
  • You have literally given no argument at all to support your claim that $(P_n(x))$ does not converge to any polynomial. – Eric Wofsey Feb 19 '19 at 22:46
  • @EricWofsey Yes, I understood that. It can still be proven that it is not complete under any sub-multiplicative norm using this argument as a partial answer. Anyway, the problem here seems to be that $|x^k|$ might become small very fast, making the series converge. So, if I can control the size of my basis, the problem will disappear. Do you agree? – stressed out Feb 19 '19 at 22:49
  • I don't know what you think the size of $|x^k|$ has to do with it. – Eric Wofsey Feb 19 '19 at 22:53
  • @EricWofsey Because if $|x^k|$ doesn't become very large, the terms of the sequence won't get close to $0$ after a certain degree. Hence, it won't have a finite degree. Am I right? – stressed out Feb 19 '19 at 22:55
  • What "won't have a finite degree"? Note that if $(P_n(x))$ converges to $P(x)$, the coefficients of $P(x)$ do NOT have to be the limit of the coefficients of the $P_n$. – Eric Wofsey Feb 19 '19 at 22:56
  • @EricWofsey Hmm. That's a good point. Thanks. Why the coefficients of $P(x)$ do not have to be the limit of the coefficients of the $P_n$'s? – stressed out Feb 19 '19 at 22:58
  • Well, why would they have to be? There is absolutely no reason that a functional taking a polynomial to one of its coefficients should be continuous with respect to some arbitrary norm. – Eric Wofsey Feb 19 '19 at 23:00
  • @EricWofsey Hmm. Because $$|(a_0+a_1x+a_2x^2+\cdots)-(b_0+b_1x+b_2x^2+\cdots)| \leq |a_0-b_0||1|+|a_1-b_1||x|+\cdots$$ So, if $|x^k|$ is finite for all $k$, then it's true. No? Or am I proving continuity for another problem? (I'm thinking now) – stressed out Feb 19 '19 at 23:05
  • What, exactly, are you claiming that that inequality proves? The left side could be small even if the right side is large. – Eric Wofsey Feb 19 '19 at 23:09
  • @EricWofsey Yes, I was thinking that if the coefficients converge to each other, then the polynomials converge. But you seem to be saying the exact opposite: if the polynomials converge to each other, the coefficients don't need to converge to each other. So, I need to prove the continuity of the projection map to disprove it. Right? – stressed out Feb 19 '19 at 23:10
  • One thing to keep in mind is that all linearly independent sets are "the same" if yoou have no structure besides a vector space structure. Given any sequence $(v_n)$ of linearly independent vectors in any vector space $V$, there is a linear injection $\mathbb{P}\to V$ sending $x^n$ to $v_n$. So if you are claiming something is true of the $x^n$ for an arbitrary norm on $\mathbb{P}$, you are also claiming the same thing is true for any sequence of linearly independent vectors in any normed vector space. – Eric Wofsey Feb 19 '19 at 23:12
  • Pretty much all the claims you've been making can easily be disproved by picking some simple sequence of linearly independent vectors in your favorite Banach space, e.g. in $\ell^2$. – Eric Wofsey Feb 19 '19 at 23:13
  • @EricWofsey Yes. Now I'm seeing the subtlety. I can cheat and resort to Baire's category theorem, but I really like this answer and I'm trying to save it from death. :( What is an example of a norm that makes the above series converge? Can you think of any now? – stressed out Feb 19 '19 at 23:13
  • It's a little messy to write down an explicit example, but you can easily find a sequence of linearly independent unit vectors $(v_n)$ in $\ell^2$ such that $\sum v_n/2^n=v_0/2$, say. Let $v_1$ be close to $-v_0$ (but perturbed slightly to be linearly independent) and then choose the $v_n$ for $n>1$ to cancel out the error. – Eric Wofsey Feb 19 '19 at 23:20
  • @EricWofsey That makes sense. Then Baire's category theorem it is. Right? – stressed out Feb 19 '19 at 23:23
  • Guys: to remove the confusion and condition of being over stressed, I simplified the problem; yet the simple answer is Taylor series expansion of exponential function as @b00n heT said before. – Saeed Feb 20 '19 at 23:53
  • @Rob Arthan: The original question comes from a textbook about Robust Control Theory – Saeed Feb 20 '19 at 23:54
  • @stress Out: thanks for thorough answer. – Saeed Feb 20 '19 at 23:55
  • @Saeed You're welcome. – stressed out Feb 20 '19 at 23:56