Let´s take the unit interval $[0,1]$ and space of polynomials $P={ c_n x^n, x \in [0,1]}$ this is a metric space with the integral metric $\rho(A,B)= \int\limits_0^1 \vert A(x) – B(x) \vert {\rm d}x $ for $A,B \in P$.
I believe this is not a complete space due to various sequences which converge to functions instead of polynomials via the Taylor expansion.
Concretely take $f_n(x)= \sum_{k=0}^n \frac{x^k}{k!}$, now it is obvious this sequence converges to $e^x$ which is not a polynomial, however, I am having problems proving this sequence is Cauchy in the integral metric.
My attempt so far is the following: without the loss of generality take $m>n$ and
$$
\rho(f_m,f_n) = \int \limits_0^1 \vert \frac{x^m}{m!}+\frac{x^{m-1}}{(m-1)!}+\cdots+\frac{x^{n+1}}{(n+1)!} \vert {\rm d}x
=
\frac{1}{(m+1)!}+\cdots+\frac{1}{(n+2)!}
=
\sum_{l=n+2}^{m+1}\frac{1}{l!}
$$
Where we can ignore the absolute value since all terms are positive on the interval $[0,1]$.
Now it is clear each term in the sum gets smaller as $m,n$ increases but also notably the amount of terms increases as $m$ increases if we fix $n$. For this sequence to be Cauchy, we need to find $n_0$ such that $\rho(f_m,f_n)<\epsilon$ for all $m,n>0$ and all $\epsilon>0$. In particular I can take $n=n_0$ and $m=n_0+k$ and it follows that $\rho(f_{n_0+k},f_{n_0})$ must go to zero as $k \rightarrow \infty$ is this correct? But I actually get it goes to infinity if I try to compute the limit, I do the following:
$$
\frac{1}{(m+1)!}+\cdots+\frac{1}{(n+2)!}
=
\frac{1+(m+1)+(m+1)m+\cdots+(m+1)m\cdots(n+3)}{(m+1)!}
\leq
\frac{1+(m+1)+(m+1)^2+\cdots+(m+1)^{m-n-1}}{(m+1)!}
=
-\frac{1}{m(m+1)!}\left(1-(m+1)^{m-n}\right)
$$
By summing the geometric series. Now I can plug in $n=n_0$, $m=n_0+k$ and take the $k\rightarrow \infty$ limit, but putting this limit into wolfram Mathematica it says it's infinite, which I agree with if I use the Stirling approximation for the factorial.
Did I enlarge the sum too much in the second step? Or is there some fault in my argument? Is there some better way of proving this sequence is Cauchy in the given metric?
Best Answer
Note that$$\sum_{l=n+2}^{m+1}\frac1{l!}<\sum_{l=n+2}^\infty\frac1{l!}.$$On the other hand,$$\lim_{n\to\infty}\sum_{l=n+2}^\infty\frac1{l!}=\lim_{n\to\infty}\left(e-\sum_{l=0}^{n+1}\frac1{l!}\right)=0,$$and therefore, given $\varepsilon>0$, there is some $N\in\Bbb N$ such that$$n\geqslant N\implies\sum_{l=n+2}^\infty\frac1{l!}<\varepsilon.$$So,$$m,n\geqslant N\implies\rho(f_m,f_n)<\varepsilon.$$