Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C=C[0,1]$ be the space of all continuous functions on $[0,1]$. Define $\|f \|=\max \ |f(x)|$. I want to show that $C$ is a Banach space.

Below is my attempt and I was wondering if it's ok.

I know I have to show that $C$ is a complete normed space.
Clearly, $\|f\| \geqslant 0$ and $\|f\|=0 \Leftrightarrow f=0$. $\|cf \|=\max~|cf(x)|=|c|\max |f(x)|=|c| \cdot \|f\|$.
$\|f+g\|=\max~|f(x)+g(x)|\leq \max~|f(x)|+\max~|g(x)|=\|f\|+ \|g\|$.
So $C$ is a normed space.

Next, I show that every Cauchy sequence in $C$ is convergent.
Let $\{f_n\}$ be a Cauchy sequence in $C$.
Let $\varepsilon \gt 0.$ Then $\exists$ an $N_1$ such that $$ \max~|f_n(x)-f_m(x)| \lt \frac{\varepsilon}{2}$$ for $n, m \gt N_1$ and $x\in[0,1]$.

But there is a subsequence $f_{k_n} $, which converges to $f$. So $\exists$ an $N_2$ such that $$ \max~\left|f_{k_n} - f\right|\lt \frac{\varepsilon}{2}$$ for each $n\gt N_2$.

Now Let $N = \max\{N_1, N_2\}$, if $n \gt N$ then $k_n \geqslant n\gt N$. So we have $$ \max~\left|f_n(x) - f(x)\right| \leqslant \max~\left|f_n - f_{k_n}\right| + \max~\left| f_{k_n} - f\right| \lt\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ Thus, $\|f_n-f\| \to 0$ as $n\to \infty$. $\quad \square$


share|cite|improve this question
You have shown that $f_n \to f$ but you have neglected to show that $f \in C[0,1]$. – nullUser Nov 20 '11 at 4:45
@Kb100: Thanks for the reminder. So the convergence is uniform in $C[0,1]$, hence $f\in C$. Is this enough? – Nana Nov 20 '11 at 5:23
Showing arbitrary cauchy sequence in $C[a,b]$ converges in $C[a,b]$ is enough. But, how do you know that there is a subsequence of $f_{n}$ that converges? Which result do you use? Justify it please. – Hassan Muhammad Nov 20 '11 at 11:30
@Nana: The statement that a sequence of continuous functions on $[0,1]$ which converges uniformly converges to a continuous function is equivalent to the fact that $C[0,1]$ is complete, so I wouldn't use it in your proof. Convergence under the max norm is uniform convergence in disguise. – nullUser Nov 20 '11 at 16:14
up vote 12 down vote accepted

Where do you get this subsequence $\{f_{n_k}\}_k$? Since a Cauchy sequence is convergent if and only if it has a convergent subsequence, you're essentially assuming the result is true here.

Here's a proof that $C[0,1]$ is complete (and thus a Banach space):

Suppose $\{f_n\}$ is Cauchy in $C[0,1]$. We must show that $f_n$ converges in the $C[0,1]$ norm to an $f$ in $C[0,1]$.

We first identify the "natural candidate" for $f$:

Since $\{f_n\}$ is Cauchy in $C[0,1]$, it follows that $\{f_n(x)\}$ is Cauchy in $\Bbb R$ for each $x\in[0,1]$. This observation, together with the fact that $\Bbb R$ is complete, gives us the well-defined function $f:[0,1]\rightarrow\Bbb R$ whose rule is $f(x)=\lim f_n(x)$.

Since the terms of $\{f_n\}$ get uniformly close to each other, we expect $f$ to be uniformly close to $f_m$ for large $m$:

Now let $\epsilon>0$ and choose $M$ so that $\|f_n-f_m\|_{C[0,1]}<\epsilon$ for $n, m\ge M$. Then for each $m>M$ and for any $x\in[0,1]$: $$ \tag{1} |f(x)-f_m(x)|=\lim_{n\rightarrow\infty}|f_n(x)-f_m(x)|\le \lim_{n\rightarrow\infty}\|f_m-f_n \|_{C[0,1]}\le\epsilon. $$

And, we finish up with some hand waving that should not seem arcane to someone studying Banach spaces:

From $(1)$, it follows that $f_n$ converges uniformly to $f$ on $[0,1]$. From this, it follows that $f\in C[0,1]$ (a uniform limit of continuous functions is continuous) and that $f_n$ converges to $f$ in $C[0,1]$.

Edit: A comment above leads me to remark:

$f$ is indeed continuous: Given $x\in[0,1]$ and $\epsilon>0$, choose $m$ so that $||f_n-f\,||_{C[0,1]}<\epsilon/3$ for $n\ge m$ and choose $\delta>0$ such that $|f_m(x)-f_m(y)|\le \epsilon/3$ for all $y$. Then if $|x-y|<\delta$: $$ |f(x)-f(y)| \le|f(x)-f_m(x)|+|f_m(x)-f_m(y)|+|f_m(y)-f(y)|<\epsilon. $$

share|cite|improve this answer
Hello, I know this was answered a long time ago, but I just want to ask why the uniform convergence of $f_n$ imply convergence of $f_n$ with respect to the maximum norm? – Kurome May 14 at 11:36
@Kurome Uniform convergence means that given $\epsilon>0$, for sufficiently large $n$ and for all $x\in[0,1]$, $|f_n(x)-f(x)|<\epsilon$. So, then, $\sup_{x\in[0,1]} |f_n(x)-f(x)|\le \epsilon$ for sufficiently large $n$. This means we have convergence in $C[0,1]$. – David Mitra May 14 at 11:41
Ah yes, that was straightforward. I just want to be clear. Let's say the maximum of the absolute difference of $f_n$ and $f$ is at $x$, then the same $N$ works so that $|f_n(x)-f(x)|<\epsilon$. – Kurome May 14 at 11:44
@Kurome Yes, that's right. – David Mitra May 14 at 11:50
Ok last question, doesn't this proof also work for the more general case where $[a,b]$ is a closed and bounded interval? – Kurome May 14 at 12:10

Let $f_{n}$ be arbitrary Cauchy sequence in $C[0,1]$.

Then for a fixed $t\in C[0,1]$ $$|f_{n}(t)-f_{m}(t)|< \epsilon \text{ for all $m,n>N$ a natural number}$$

That means $f_{n}$ is a Cauchy sequence in the set of real numbers. And since the set of reals is complete, there exist $f(t_{0})\in \Bbb R$ such that $f_{n}\to f(t_{0})$ as $n\to \infty$ with $t_{0}$ arbitrary in $C[0,1]$.

For $m\ge n$, and allowing n to go to infinity, we have $$\max_{t \in{[0,1]}}|f(t)-f_{m}(t)|<\epsilon$$ $$\implies \|f-f_{m}\|<\epsilon$$

For all $n$ bigger than a natural number $N$.

Thus, $f_{n}\to f$ as $n \to \infty$.

From here you can use the uniform convergence to show that $f$ is in $C[0,1]$

share|cite|improve this answer
Is your $t_0$ supposed to be the same as $t$ which you fixed before? Did you want to write that $f_n(t)$ is a Cauchy sequence in $\mathbb R$ and in the next line $f_n(t_0) \to f(t_0)$ instead of $f_n \to f(t_0)$? – Martin Sleziak Nov 20 '11 at 12:22

The usual way to prove that $X$ compact Hausdorff $\implies C(X,\mathbb K)$ is a Banach space over $\mathbb K$ goes in two steps:

  1. Show that $B(X) := \ell^\infty(X,\mathbb K) = \{f \in \mathbb K^X \mid f$ is bounded $\}$ is a Banach space w.r.t. the sup-norm.
  2. Show that the uniform limit of a sequence of functions that are continuous at a point is continuous at that point as well. From this it follows that $C(X)$ is closed in $B(X)$.
share|cite|improve this answer
You don't need Hausdorff. And, taking $C$ to mean bounded continuous functions, you don't need compact either. – Rasmus Nov 21 '11 at 20:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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