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 $\mathbb{P}$?


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    $\begingroup$ One important note : With respect to what norm ? $\endgroup$ – Rebellos Feb 19 at 20:28
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    $\begingroup$ wrt any norm in vector space $\endgroup$ – Saeed Feb 19 at 20:30
  • $\begingroup$ I think that one counterexample which may interest you is the sequence of polynomials which defines the Taylor series for $e^x$ $\endgroup$ – b00n heT Feb 19 at 20:33
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    $\begingroup$ @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? $\endgroup$ – Ingix Feb 19 at 20:45
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    $\begingroup$ 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. $\endgroup$ – Rob Arthan Feb 19 at 20:48

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.

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    $\begingroup$ +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. $\endgroup$ – Rob Arthan Feb 19 at 21:09
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    $\begingroup$ Nice solution, and the fix was 'simple'! Upvoted, and I'll delete my comments, as they are irrelevant now. $\endgroup$ – Ingix Feb 19 at 22:09
  • $\begingroup$ @Ingix Thank you. Your comment was crucial in improving the answer. I appreciate it. $\endgroup$ – stressed out Feb 19 at 22:11
  • $\begingroup$ This is wrong: you have never proved that the sequence does not converge, and in fact it does converge for some norms. $\endgroup$ – Eric Wofsey Feb 19 at 22:44
  • $\begingroup$ @EricWofsey Do you mean that the speed of $\|x^k\|$ becoming small might make it converge? $\endgroup$ – stressed out Feb 19 at 22:45

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