The normed vector space of continuous function is complete

Question

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

Answer 1

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 $m>N_\varepsilon$.

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