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Question: Let $P[0,1]$ be the set of all polynomials defined on $[0,1]$. A metric is defined $P[0,1]$ by $d(P_1, P_2)=\sup\limits_{0\leq x\leq 1}|P_1(x)-P_2(x)|$. Then show that this metric space is incomplete.

We know that a metric space is complete if every Cauchy sequence in the metric space is convergent in that metric space. I cannot understand what will be the approach? Here every element in $P[0,1]$ is polynomial. How can I get sequence from $P[0,1]$.

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    $\begingroup$ Consider $e^x$ on $[0,1]$. Why is it not a polynomial? Does it have a series expansion? Can you find polynomials going to $e^x$ uniformly? $\endgroup$ – астон вілла олоф мэллбэрг Feb 9 at 6:06
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Take any real analytic function (function which equals its power series), but which is not a polynomial. For instance $\ln (1+x),e^x,\sin x,\cos x$ etc.

Then it can be approximated by polynomials.

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  • $\begingroup$ My mistake. Thanks. $\endgroup$ – Chris Custer Feb 9 at 6:59
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So, what happens if we try to find the limit that should be there? Well, from the definition of that distance, $\left|P_1(x)-P_2(x)\right| \le d(P_1,P_2)$ for any fixed $x$. As such, if a sequence $Q_n$ is Cauchy, its values $Q_n(x)$ at a particular point are also a Cauchy sequence - in $\mathbb{R}$. That sequence has a limit, because $\mathbb{R}$ is complete.

So then, we have a candidate limit: given a Cauchy sequence $Q_n$, the function $$Q(x)=\lim_{n\to\infty} Q_n(x)$$ "should" be the limit of the $Q_n$. What could go wrong? Two possibilities:

  • That's a pointwise limit, not directly using the metric. What if it's not convergent in the metric sense - that $d(Q_n,Q)\neq 0$?
  • What if $Q$ isn't a polynomial? We defined it as a function, after all.

As it turns out, the first issue isn't a problem - we indeed have $d(Q_n,Q)\to 0$. The second, though? There are lots of ways to get a sequence of polynomials that converges to something that's not a polynomial.

For an example, how about a geometric series? Consider the sequence $$Q_n(x) = 1+\frac{x}{2}+\frac{x^2}{4}+\cdots+\frac{x^n}{2^n} = \sum_{k=0}^n \left(\frac x2\right)^n = \frac{1-\frac{x^{n+1}}{2^{n+1}}}{1-\frac x2}$$ Since $\frac{x^{n+1}}{2^{n+1}} \le \frac1{2^{n+1}}\to 0$, we have $Q_n(x)\to Q(x)=\frac{1}{1-\frac x2}=\frac{2}{2-x}$, with error $\sup_x |Q_n(x)-Q(x)| \le 2^{-n}$ and $d(Q_n,Q_m)\le 2^{-\min(m,n)}$. That's a Cauchy sequence that doesn't converge in $P[0,1]$, because its limit isn't a polynomial.

As a side note, there's a theorem: for any continuous function $f$ on the closed interval, there is a sequence of polynomials $f_n$ that converges uniformly (that is, $\lim_n d(f_n,f) = 0$ in the metric here) to $f$. We can get basically everything.

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