# Show that $C[a,b]$ is a complete space under the metric $d(f,g)=\sup_{t\in [a,b]}|f(t)-g(t)|$.

$C[a,b]$ is a normed vector space of all continuous complex valued functions on $[a,b]$, with supremum norm

$$\|f\|_\infty=\sup_{t\in [a,b]}|f(t)|.$$

The metric induced by the norm is $$d(f,g)=\|f-g\|_\infty = \sup_{t\in [a,b]}|f(t)-g(t)|.$$

Show that $C[a,b]$ is a complete metric space under this metric induced by the supremum norm (i.e. show $C[a,b]$ is a Banach space).

• HINT: Set up a Cauchy sequence under that norm, then try to find where the limit is, see if that limit is continuous and has a finite supremum. – Shuhao Cao Oct 28 '13 at 1:59

If $\langle f_n\rangle$ is Cauchy then for any $\varepsilon >0$ there exists $N$ such that $n,m\geqslant N$ gives $$\sup_{x\in[a,b]}|f_n(x)-f_m(x)|<\varepsilon$$
But then the sequence of complex numbers $\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges. This means we can define $f:[a,b]\to \Bbb C$ by $$f(x)=\lim\limits_{n\to\infty}f_n(x)$$
It remains to show that $f_n\to f$ uniformly, and that $f$ is continuous. Can you do this?
• "$\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges." But isn't this what we have to show? – MathsMy Oct 28 '13 at 2:18
• Oh, if $f_n \rightarrow f$ uniformly then $f$ is continuous right? – MathsMy Oct 28 '13 at 2:19
• In Complex numbers, convergence $\iff$ Cauchy. But in other sets, isn't it Convergence $\rightarrow$ Cauchy but not the other way? – MathsMy Oct 28 '13 at 2:25
• @Dosomemaths Sure. But you're assuming your function goes into $\Bbb C$. – Pedro Tamaroff Oct 28 '13 at 2:26