# The normed vector space of continuous function is complete

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

• What is the usual distance metric? – NickD May 5 '17 at 19:31
• $d(x,y) = |x-y|$, I thought it's standard notation – asdf May 5 '17 at 19:41
• But what does that mean in the context of the space you are considering? Does it make sense? – NickD May 5 '17 at 20:00
• 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? – asdf May 5 '17 at 20:08
• I guess the question is: what is X? You have to be able to say that two elements of X are "close" somehow. – 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.

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

• Yes, I am OK with that bit, what confuses me is the continuity part – asdf May 5 '17 at 21:48
• @asdf Just a standard trick – egreg May 5 '17 at 21:59
• 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 – asdf May 5 '17 at 22:07
• @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. – 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.

Q.E.D.