First of all, let $(f_n)$ be a Cauchy sequence in $B(x)$ which is the vector space of bounded functions $f\colon X \to \mathbb R$ equipped with the norm $\|f\| = \sup|f(x)|$.

Note that $|f_n(x)-f_m(x)| \leq \|f_n-f_m\|$ which tends to $0$, therefore $f_n$ is a Cauchy sequence in $\Re$, hence we can define a function

$f\colon X \to \mathbb R$ s.t $f(x)= \lim f_n(x)$.

Now we have to show that $f$ is in $B(X)$, $f_n \to f$ in $B(X)$ and that $f$ is continuous.

I am OK with the proofs for the first $2$. For the last one - do I have to show that it is continuous wrt the norm, or with respect to the usual distance metric?

Edit: $(X,d)$ is a metric space

  • $\begingroup$ What is the usual distance metric? $\endgroup$ – NickD May 5 '17 at 19:31
  • $\begingroup$ $d(x,y) = |x-y|$, I thought it's standard notation $\endgroup$ – asdf May 5 '17 at 19:41
  • $\begingroup$ But what does that mean in the context of the space you are considering? Does it make sense? $\endgroup$ – NickD May 5 '17 at 20:00
  • $\begingroup$ As $C_b(X)$ consists of the real bounded continuous functions, if we prove that it is continuous in the reals then we'll be done? $\endgroup$ – asdf May 5 '17 at 20:08
  • $\begingroup$ I guess the question is: what is X? You have to be able to say that two elements of X are "close" somehow. $\endgroup$ – NickD May 5 '17 at 20:29

The vector space $B(X)$ of bounded functions $f\colon X\to\mathbb{R}$ makes sense with no further structure on $X$.

What you have to show is that, given a Cauchy sequence $(f_n)$ in $B(X)$ there exists a bounded function $f\in B(X)$ such that $$ \lim_{n\to\infty}f_n=f $$ First of all, we can identify a function that should be the limit (provided it is bounded). Indeed, if $x\in X$, the sequence $(f_n(x))$ in $\mathbb{R}$ is Cauchy (easy proof), so we can define $$ f(x)=\lim_{n\to\infty}f_n(x) $$ Note that if the sequence converges to some function $g$ in $B(X)$, then it must be $\lim_{n\to\infty}f_n(x)=g(x)$, for every $x\in X$. Thus only the “pointwise limit” above can work.

Our tasks now are

  1. to prove that $f$ is bounded, and that
  2. $\lim\limits_{n\to\infty}f_n=f$.

Let's try point 1. For every $\varepsilon>0$, there exists $N_\varepsilon$ such that, for $m,n>N$, $$ \|f_n-f_m\|<\varepsilon $$ This implies, for all $n>N$, fixing any $m>N$, $$ f_m(x)-\varepsilon<f_n(x)<f_m(x)+\varepsilon \qquad\text{(for all $x\in X$)} $$ Passing to the limit for $n\to\infty$, we get $$ f_m(x)-\varepsilon<f(x)<f_m(x)+\varepsilon \tag{*} $$ so $f$ is indeed bounded (fill in the details).

On the other hand, (*) holds for every $m>N_\varepsilon$, so $$ |f(x)-f_m(x)|<\varepsilon $$ for every $x\in X$ and so $\|f-f_m\|<\varepsilon$, for every $m>N_\varepsilon$.

Now, suppose all $f_n$ are continuous; let $x\in X$ and fix $\varepsilon>0$. There is $N$ such that, for $n>N$, $\|f-f_n\|<\varepsilon/3$, for every $n>N$. Choose one $n>N$; then there is $\delta>0$ such that $d(x,y)<\delta$ implies $$ |f_n(x)-f_n(y)|<\varepsilon/3 $$ Then, if $d(x,y)<\delta$, we have $$ |f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|<\varepsilon $$

  • $\begingroup$ Yes, I am OK with that bit, what confuses me is the continuity part $\endgroup$ – asdf May 5 '17 at 21:48
  • $\begingroup$ @asdf Just a standard trick $\endgroup$ – egreg May 5 '17 at 21:59
  • $\begingroup$ OK, but do we have to prove that the function is continuous under the standard metric, i.e. to prove that $f$ is continuous as a function between $X$ and the reals or under the metric on $B(X)$. I am OK with the former, but the latter confuses me. Thanks $\endgroup$ – asdf May 5 '17 at 22:07
  • $\begingroup$ @asdf $f$ is a function from $X$ to $\mathbb{R}$. It doesn't make sense to say it's continuous with respect to the metric on $B(X)$: it would be like saying a point in $\mathbb{R}$ is continuous. $\endgroup$ – egreg May 5 '17 at 22:12

To prove the completeness, we need to prove for any Cauchy sequence $f_1,f_2,\dotsm \in E$ satisfying that

$\forall \epsilon >0$, there exists a number $n_0$ such that for all $m,n\ge n_0$ $$ ||f_m-f_n||<\epsilon\\ \max_{x\in[a,b]} |f_m(x)-f_n(x)|<\epsilon $$ that implies $$ |f_m(x)-f_n(x)|<\epsilon \quad \forall n,m\ge n_0,\forall x\in[a,b] $$ that means $(f_n(x))$ is Cauchy sequence $\forall x\in[a,b]$.

We denote $$ f(x)=\lim_{n\rightarrow \infty } f_n(x),\forall x\in[a,b] $$ Let $n\rightarrow \infty $ , we have $$ |f_m(x)-f(x)|<\epsilon,\forall m\ge n_0 ,\forall x\in[a,b]\\ \max_{x\in[a,b]}|f_m(x)-f(x)|<\epsilon \\ ||f_m-f|<\epsilon $$ Then we prove $f\in E$

Let $\forall x_0\in[a,b]$. Since $f_{n_0}$ is continuous on $[a,b]$, there exists $\delta >0$ such that $|f_{n_0}(x_0)-f_{n_0}(y)|<\epsilon$ for every $y\in[a,b]$ and $|x_0-y|<\delta$ $$ |f(x_0)-f(y)|= |f(x_0)-f_{n_0}(x_0)+f_{n_0}(x_0)-f_{n_0}(y)+f_{n_0}(y)-f(y)|\\ \le |f(x_0)-f_{n_0}(x_0)|+|f_{n_0}(x_0)-f_{n_0}(y)|+|f_{n_0}(y)-f(y)|\\ < 3\epsilon $$ whenever $|x_0-y|<\delta$, that means $f$ is continuous.



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