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Let $P$ be the set of all polynomials and consider the sup-norm on $[0,1]$ by $$||p||_\infty = \sup\{|p(x)| : x \in [0,1]\}.$$ I need to show $P$ is not complete in the sup-norm on the interval $[0,1]$. To do this, it suffices to show there exists a sequence of elements in $P$ which is Cauchy but has no limit in $P$. To this end, consider the sequence of polynomials $(p_n)$ defined by $$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k},$$ where $x \in [0,1]$. I have already proved that $(p_n)$ is in fact Cauchy with respect to the sup-norm, but I am confused on how to show that $$p(x) = \frac{1}{1 - \frac{x}{2}}$$ is the only possible limit for the sequence $(p_n)$. We know that $$\lim_{n \to \infty} p_n(x) = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{x^k}{2^k} = \sum_{k=0}^{\infty} \left(\frac{x}{2}\right)^k = \frac{1}{1 - \frac{x}{2}} = p(x)$$ for $x \in [0,1]$. Also, $\sum_{k=0}^\infty \left(\frac{x}{2}\right)^k = \sum_{k=0}^\infty \frac{1}{2^k} x^k$, is a power series with radius of convergence $(-2,2).$ Therefore, since $[0,1] \subseteq (-2,2)$, it follows that $\sum_{k=0}^\infty \left(\frac{x}{2}\right)^k = \sum_{k=0}^\infty \frac{1}{2^k} x^k$ converges uniformly on $[0,1]$ and is continuous on $[0,1]$. I believe this would then imply that $p(x)$ is in the larger space $C([0,1])$ and hence, $p(x)$ is the only possible limit for our Cauchy sequence $(p_n)$. Is this argument valid? Do I even need to state any of this? Thanks.

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Your initial example is correct.

If a sequence of functions converges uniformly, it converges pointwise. And for each point $x \in [0,1]$, $(p_n(x))$ converges to $p(x)$. Hence if $(p_n)$ converges for the infinite norm, it can only be towards $p$.

To conclude, it remains to be proven that $p$ is not a polynomial. This is not difficult as for any integer $n$, $p^{(n)}$ is not always vanishing.

By the way a theorem states that a normed vector space with a countable basis is never complete.

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  • $\begingroup$ Sorry for the late response, I was called into work. Ok, so since $(p_n)$ is Cauchy in the sup-norm, does that imply that $(p_n) \rightarrow p(x)$ in the sup-norm? If it does, then we can further say that this convergence is uniform on $[0,1]$. Therefore, following what you said, $p(x)$ can only be the possible limit candidate. $\endgroup$ – ATLKing Mar 11 '17 at 18:34
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Indeed you are trying to show that set of all polynomials complete! you have to show that the limit of a series of polynomial is not a polynomial. One of the ways is proving by counterexample. We have $$\lim_{n\rightarrow \infty} 1-x^2/2+x^4/4-\cdots=cosx$$ which is not polynomial then the space is not complete.

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