# Convergence with Supremum Metric

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!

• Show that $f_n$ is not Cauchy. – copper.hat Oct 9 '18 at 18:30
• 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 – qbert Oct 9 '18 at 18:39
• 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? – Albert B Oct 9 '18 at 18:42
• @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. – 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.