2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $$ |f(x)-f(y)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|<\varepsilon $$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

Q.E.D.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.