Let B[0,1] be the set of bounded functions on [0,1 equipped with the supremum metric. Let ($f_n$) be the sequence in B[0,1] defined by,

$f_n(x)$ = $\Bigg\{$ 1 - $nx$ if 0 $\leq$ $x$ $\leq$ $\frac 1n $ or 0 if $\frac 1n$$\leq$ $x$ $\leq$ 1

Prove that ($f_n$) does not converge in {$B$[0, 1], d}.

After drawing what the graph of the function looks like, I can get a picture of it in my head but I am unsure of what steps to take to prove it. Help would be much appreciated!

  • $\begingroup$ Show that $f_n$ is not Cauchy. $\endgroup$ – copper.hat Oct 9 '18 at 18:30
  • $\begingroup$ Alternatively, and with more machinery, note that the $f_n$ are continuous, while their pointwise limit is not. The uniform limit of continuous functions is continuous $\endgroup$ – qbert Oct 9 '18 at 18:39
  • $\begingroup$ I have that d($f_n$,$f_m$) = 1 - $\frac mn$ for any n greater or equal to m, does that prove that it is not cauchy? $\endgroup$ – Albert B Oct 9 '18 at 18:42
  • $\begingroup$ @AlbertB: Yes, because for any $n$ you have $d(f_{2n},f_n) = {1 \over 2}$. Hence $f_n$ is not Cauchy. However, qbert's suggestion has more intuition. $\endgroup$ – copper.hat Oct 9 '18 at 19:02

Observe that $f_n(x) \to 0 $ as $n \to \infty$ for any $x\in (0,1]$ and $f_n(0) = 1$ for all $n\in \mathbb{N}$. Hence, if $\{f_n\}$ converges to some function $f\in B[0,1]$ in sup. metric, it has to be the function $$f_0(x) = \begin{cases} 1, \text{ if } x = 0 ,\\ 0, \text{ if } x\in(0,1] \end{cases}$$ since convergence in supremum metric implies point-wise convergence. But then $$ \sup\limits_{x\in [0,1]} |f_n(x) - f_0(x) | \geq |f_n(\frac{1}{2n})| = \frac 12 , $$ which contradicts to the assumption that $\{f_n\}$ converges to $f_0$ in supremum metric.


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