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 $$f_n(x) = \begin{cases} 1-nx, & 0 \leq x \leq \frac{1}{n} \\ 0, & \frac{1}{n} \leq x \leq 1. \end{cases}$$ Determine whether $(f_n)$ is a Cauchy sequence in $B[0,1]$.

If $(f_n)$ is a Cauchy sequence, then $\forall \epsilon > 0 \ \exists N \in \mathbb{N}$ such that $$m,n > N \implies \sup_{x \in [0,1]} \left| f_n - f_m \right| < \epsilon.$$ Observe that for $0 \leq x \leq 1/n$, \begin{eqnarray*} \sup_{x\in [0,1]} \left| f_n(x) - f_m(x) \right| &=& \sup_{x \in [0,1]} \left| (1-nx) - (1-mx) \right| \\ &=& \sup_{x \in [0,1]} \left| mx - nx \right| \\ &=& m -n. \end{eqnarray*}

Do we therefore conclude that $(f_n)$ is Cauchy in $B[0,1]$? From the above argument, we see that the distance between $f_n$ and $f_m$ depends only on the difference between the indices.

Also, part (b) of this questions asks us to show that $f_n$ does not converge. However, by drawing a simply sketch, we see that $f_n \to f$, where $$f(x) = \begin{cases} 1, & \text{if} \ x =0 \\ 0, & 0 < x \leq 1. \end{cases}$$ This is a bounded function, and so, $f \in B[0,1]$. Is there something wrong with the question or am I missing something?


Think of it intuitively, by definition all $f_n$ are between zero and one, yet your argument lets you get $|f_n-f_m|=m-n$ which could be as large as we like. The problem is you taking the supremum over $x\in[0,1]$, when $f_n$ is zero outside of $[0,\frac{1}{n})$. Anyway if your argument was correct it would imply that the sequence is not Cauchy, as $m-n$ can be as large as you like for $m,n\ge N$.

To solve this, without loss of generality suppose $n>m$. Then you have $$\sup_{x\in[0,1]}|f_n(x)-f_m(x)|=\mathrm{max}\left(\sup_{x\in\left[0,\frac{1}{n}\right]}\left|(1-nx)-(1-mx)\right|,\sup_{x\in\left[\frac{1}{n},\frac{1}{m}\right]}\left|0-(1-mx)\right|\right).$$

You should be able to show that this is equal to $$=1-\frac{m}{n},$$ where again remember $n>m$ (drawing a picture might help).

To prove that a sequence is Cauchy, you need to show that for every $\epsilon>0$, there exists an $N>0$ such that $m,n>N$ implies $\sup_{x\in[0,1]}|f_n-f_m|<\epsilon$. Suppose you are given some $N$, and consider $n\ge m\ge N$. How large can you make $\sup|f_n-f_m|=1-\frac{m}{n}$? In this case, can the sequence be Cauchy? Hint: first consider maximising this with respect to $m$, then $n$. It might help to draw this graphically, fixing $m$ and seeing what you can do with $n$.

For part b, when you say a sequence converges, you need to say in which sense. Clearly $f_n\rightarrow f$ pointwise, but that does not necessarily mean it converges in the supremum norm, i.e. that $\sup_{x\in[0,1]}|f_n-f|\rightarrow 0$ (for an example of this see Pointwise but not Uniformly Convergent). You can either use the result of Part a, or think about the fact that each $f_n$ is continuous, and so must take each value between $0$ and $1$.

  • $\begingroup$ Thanks, I understand your supremum correction (thanks a lot for that, I'm glad you pointed that out), however, how do I go from that to "... therefore $(f_n)$ is not a Cauchy sequence"? $\endgroup$ – user319128 Jun 12 '16 at 1:31
  • $\begingroup$ No worries. I have updated my answer, does it make sense? It took me a little while to think about it, drawing some pictures might help. $\endgroup$ – Ruvi Lecamwasam Jun 12 '16 at 1:54

Hint: $\|f_{2n}-f_n\|_{\infty}=sup_{x\in [0,1]}\mid f_{2n}(x)-f_n(x)\mid$.

You have $\mid f_{2n}({1\over 2n})-f_n({1\over 2n})\mid=\mid f_n({1\over 2n})\mid =\mid 1-{n\over 2n}\mid={1\over 2}$.

Deduce that $(f_n)$ is not a Cauchy sequence.

  • $\begingroup$ $f_n$ converges simply i.e $lim_n f_n(x)=0$, but $f_n$ does not converge uniformly. Why my argument is not correct? $\endgroup$ – Tsemo Aristide Jun 12 '16 at 1:12
  • $\begingroup$ My bad, you are right. :) I forgot checking the uniform convergence. $\endgroup$ – Zhanxiong Jun 12 '16 at 1:17

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