# Is $C(\mathbb R)$ Complete?

I'm trying to prove an exercise from Carthers' book chapter10 of Real Analysis, problem claimed as,

where $C(\mathbb R)$ denote the infinity norm space of all continuous functions on real line.

I tried to use the hint. However, I got an counterexample that probably works for incomplete $C([-n,n])$. That is: $f_n(x) = x^{2n}$ does not converge to a continuous function in $C([-1,1])$. Where is my fault?

• Your sequence is not Cauchy.
– Pedro
Commented Feb 7, 2015 at 0:56
• @PedroTamaroff: Seems u r right? Commented Feb 7, 2015 at 1:00
• You are forgetting the metric used is the one induced by the supremum norm.
– Pedro
Commented Feb 7, 2015 at 1:06
• Are you copying your entire homework assignment, one at a time, here? Commented Feb 7, 2015 at 1:25
• @GEdgar: Seriously, no. I learned by myself, just interested in. That's it. Commented Feb 7, 2015 at 1:28

Let $$(f_n)$$ be a Cauchy sequence in $$C(\Bbb R)$$, and define $$\Phi : [0, \infty) \to \Bbb R$$ by the equation $$\Phi(u) = \frac{u}{1+u}$$. Given $$\epsilon > 0$$, there exists a positive integer $$N$$ such that $$\|f_n - f_m\| < \epsilon$$ for all $$n, m \ge N$$. Thus for every $$k$$, $$\Phi(\|f_n - f_m\|_{C[-k,k]}) < \epsilon$$ for all $$n, m > N$$. Since $$\Phi$$ is strictly increasing, then for each $$k$$, $$\|f_n - f_m\|_{C[-k,k]} < \epsilon$$ for all $$n, m \ge N$$. Therefore $$(f_n)$$ is Cauchy in $$C[-k,k]$$ for all $$k$$. Since $$C[-k,k]$$ is complete, there is a $$g_k \in C[-k,k]$$ such that $$f_m \to g_k$$ uniformly on $$C[-k,k]$$. The $$g_k$$ define a continuous function $$g\in C(\Bbb R)$$.

Choose a positive integer $$n_0$$ such that $$\sum_{k=n_0+1}^\infty \frac{1}{2^k} < \epsilon$$. The function $$\Phi$$ is bounded by $$1$$ so that $$\|f_n - g\| \le \sum_{k = 1}^{n_0} 2^{-k}\|f_n - g\|_{C[-k,k]} + \sum_{k = n_0 + 1}^\infty \frac{1}{2^k} \le \sum_{k = 1}^{n_0} 2^{-k}\|f_n - g_k\| + \epsilon$$ Since $$\|f_n - g_k\|_{C[-k,k]} \to 0$$ as $$n \to \infty$$ for every $$k$$, it follows that $$\limsup_{n \to \infty} \|f_n - g\| \le \epsilon$$ As $$\epsilon$$ is arbitrary, $$f_n \to g$$ in $$C(\Bbb R)$$.

• thanks,kobe. I've got your idea^_^ Commented Feb 7, 2015 at 3:00
• Kobe, can you explain why "fn continuous, it suffices to show that fn->f uniformly", please? Commented Feb 7, 2015 at 3:28
• It's because the uniform limit of a sequence of continuous functions is continuous.
– kobe
Commented Feb 7, 2015 at 3:29
• But that fn(x) -> f(x) while f(x) belongs to R is a pointwise converges instead of uniformly? Commented Feb 7, 2015 at 3:38
• Yes, by completeness of $\Bbb R$, I found a function $f$ that is the pointwise limit of $f$. But I went a step further to show that $f_n$ converges to $f$ uniformly on $\Bbb R$.
– kobe
Commented Feb 7, 2015 at 3:41

I believe Carothers wanted the reader to use the metric for $$C(\mathbb{R})$$ which was defined earlier in the book: $$d(f,g) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f,g)}{1+d_k(f,g)}$$ where $$d_k(f,g) = \max_{\lvert x\rvert \leq k}\lvert f(x) - g(x)\rvert$$

The proof would then go as follows:

Let $$f_n$$ be a Cauchy sequence in $$C(\mathbb{R})$$.

Now, for any $$\alpha \in \mathbb{N}$$, suppose $$f_n$$ was not Cauchy on $$[-\alpha, \alpha]$$ under the supremum norm. Then there exists an $$\epsilon > 0$$ such that for all $$N \in \mathbb{N}$$ there exists an $$n, m > N$$ such that $$\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert\geq \epsilon$$. Then we would have that with this $$\epsilon$$, for all $$N \in \mathbb{N}$$ there exists an $$n, m > N$$ such that \begin{align*}d(f_n,f_m) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f_n,f_m)}{1+d_k(f_n,f_m)} \geq 2^{-\alpha}\frac{d_{\alpha}(f_n,f_m)}{1+d_{\alpha}(f_n,f_m)} &= 2^{-\alpha}\frac{\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert}{1+\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert}\\ &\geq 2^{-\alpha}\frac{\epsilon}{1+\epsilon} > 0\end{align*}

The last inequality is due to $$\frac{x}{1+x}$$ being an increasing function.

Thus $$f_n$$ would not be Cauchy in $$C(\mathbb{R})$$. So we must have that $$f_n$$ is Cauchy on $$[-\alpha, \alpha]$$ under the supremum norm for any $$\alpha \in \mathbb{N}$$.

It can be shown that $$C[-\alpha, \alpha]$$ is complete. Since you are already working in Carothers, if you need help proving this the proof is almost identical to Lemma 10.8. Since $$C[-\alpha, \alpha]$$ is complete and $$f_n$$ is uniformly Cauchy, $$f_n$$ converges uniformly to some function $$f_{\alpha}$$ in $$C[-\alpha,\alpha]$$. Since $$f_{\alpha}$$ agrees with $$f_{\beta}$$ on $$[-\beta,\beta]$$ for every $$\beta < \alpha$$, we have that $$f = \lim_{\alpha \rightarrow \infty} f_{\alpha}$$ is a well defined continuous function on $$\mathbb{R}$$.

So, let $$\epsilon > 0$$ and choose an $$\alpha$$ so large that $$2^{-\alpha} < \frac{\epsilon}{2}$$ and a $$n$$ so large that $$\max_{\lvert x\rvert < \alpha} \lvert f_n(x) - f(x)\rvert < \frac{\epsilon}{2}$$ as $$f_n$$ converges uniformly to $$f$$ on $$[-\alpha, \alpha]$$.

Then we have that:

\begin{align*}d(f_n, f) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)} &= \sum_{k=1}^{\alpha} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)} + \sum_{k=\alpha+1}^{\infty} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)}\\ &\leq \sum_{k=1}^{\alpha}\frac{\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert}{1+\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert} + \sum_{k=\alpha+1}^{\infty}2^{-k}\\ &\leq \sum_{k=1}^{\alpha}2^{-k}\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert + 2^{-\alpha}\\ &\leq \max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert + 2^{-\alpha} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align*}

• looks good! will upvote. thanks. Commented Aug 11, 2020 at 21:35
• +1. This is better than the accepted answer. Commented Aug 12, 2020 at 1:04