# The space $C_b(\mathbb{R})$ is complete

Let $C_b(\mathbb{R})$be the space of all bounded continuous functions on $\mathbb{R}$, normed with $$\|f\|= \sup_{x\in \mathbb{R}}|f(x)|$$ Show that this space is complete.

Complete mean that all Cauchy sequences converges. So if we have an Cauchy sequence $(f_n)$, define $f(x) = \lim_{n \rightarrow \infty} f_n(x)$, we much show that $f\in C_b(\mathbb{R})$ and $\|f - f_n\| \leq \epsilon.$

How can I proceed? If I take $f(x) = f_n(x) + (f(x) - f_n(x))$ and show that the last parentheses goes too zero? But then I end up with $$|\lim_{m \rightarrow \infty} f_m(x) - f_n(x)|$$

how can I proceed from here? move out the limes? is that possible? how do one reason?

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Here is a similar post. – David Mitra Dec 19 '12 at 12:43

Sketch (for a more general result):

Step 1: Prove that $\ell_\infty(X)$, the set of bounded functions on $X$, is complete for any set $X$. Sketch/Hint: Uniformly cauchy implies pointwise cauchy which implies pointwise convergent. Then show that the pointwise limit is bounded. Then show that the uniform limit exists and equals the pointwise one.

Step 2: Prove that if $X$ is a metric space (or more generally a Hausdorff topological space) and if $(f_n)$ is a sequence of functions in $\ell_\infty(X)$ with each $f_n$ continuous at $x_0 \in X$ such that $f_n$ converges uniformly to a function $f$, then $f$ is continuous at $x_0$.

Step 3: Conclude that $C_b(X)$ is closed in $\ell_\infty(X)$ and is thus complete.

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Thomas already showed how to do the boundedness.

I will take care of that $f_n$ actually converges uniformly.

First notice that $(f_n(x))_n$ is a cauchy sequence in $\mathbb{R}$ due to the fact that $f_n$ is cauchy with respect to the sup norm. Can you prove this yourself? Hence the point wise limit exists. Let $f(x) = \lim_{n\rightarrow \infty} f_n(x)$

Now let $\epsilon > 0$ and let $x \in \mathbb{R}$. Then we have \begin{align*} \lVert f(x) - f_n(x) \rVert &= \lim_{m\rightarrow \infty} \lVert f_m(x) -f_n(x)\rVert \tag{continuity of $\lVert \cdot \rVert$}\\ &< \epsilon \tag{*} \end{align*}

(*) This is the case if we take $N$ such that $\forall m,n\geq N: \lVert f_m(x) - f_n(x) \rVert \leq \epsilon$, this $N$ exists due to the fact that $f_n$ cauchy is with respect to the sup norm. So once $m$ is larger than $N$, the norm is smaller than $\epsilon$.

Can you show yourself that $f$ is continuous, this follows from the fact that $f_n \rightarrow f$ uniformly.

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Suppose $f_n\rightarrow f$ where $f_n\in C_b(\mathbb{R})$. If $n$ is too big, you can find $\epsilon>0$ such that $$|f(x)-f_n(x)|<\epsilon,\ \forall\ x\in\mathbb{R}$$

This implies that $|f(x)|<|f_n(x)|+\epsilon$. Therefore $f\in C_b$.

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