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I am reading Kolmogorov's "Introductory Real Analysis" and I have come across this example of an incomplete metric space ($C^2_{[a,b]}$) on page 59: If $$\phi_n(t)= \left\{ \begin{array}{lcc} -1 & -1 \leq t \leq -\frac{1}{n} \\ nt & -\frac{1}{n}\leq t \leq \frac{1}{n} \\ 1 & \frac{1}{n}\leq t \leq 1 \end{array} \right.$$ then {$\phi_n(t)$} is a fundamental sequence in $C^2_{[-1,1]}$, since $\\$ $\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt \leq \frac{2}{{min\{n,n^{'}\}}}$, where $C^2_{[a,b]}$ is the metric space on the set of all functions continuous on the interval [a,b], equipped with the distance function $\rho(x,y)=(\int^a_{b}[x(t)-y(t)]^2dt)^{1/2}$. It is worth noting that 'fundamental sequence' and 'Cauchy sequence' are used interchangeably here. The problem I have is that I don't understand there this upper bound comes from and why it is chosen here. From desmos:https://www.desmos.com/calculator/mrd9tjyrup, the max distance between 2 elements only ever reaches $\frac{2}{3}$, but the given bound $\frac{2}{{min\{n,n^{'}\}}}=2$ for ${min\{n,n^{'}\}}=1$. Also, it is my understanding that $\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt = \frac{2}{{min\{n,n^{'}\}}}$ is a consequence of the limit taken as $n^{'} \rightarrow \infty.$ Any help would be much appreciated.

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  • $\begingroup$ Not that of course the limit function is discontinuous, which is why this metric space is not complete. $\endgroup$ – Math1000 Dec 23 '19 at 4:45
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Let $min(n,n')=m$ then notice $\phi_n(t)$ and $\phi_{n^{'}}(t)$ agree off of $[-\frac{1}{m},\frac{1}{m}] $ so, $\int^1_{-1}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt = \int^{\frac{1}{m}}_{-\frac{1}{m}}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt$. Now notice on this interval, there difference is bounded by 1 as both functions are always of the same sign and have modulus at most one. Also the length of the interval is $ \frac{2}{m}$ so by the max length bound we get $$ \int^{\frac{1}{m}}_{-\frac{1}{m}}[\phi_n(t)-\phi_{n^{'}}(t)]^2dt\leq \frac{2}{m} $$

So it is not a consequnce of the limit. The bound is a crude bound but most importantly it shrinks as the m goes to infinity.

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