$C[a,b]$ 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).

  • $\begingroup$ 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. $\endgroup$ – 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?

  • $\begingroup$ "$\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges." But isn't this what we have to show? $\endgroup$ – MathsMy Oct 28 '13 at 2:18
  • $\begingroup$ Oh, if $f_n \rightarrow f$ uniformly then $f$ is continuous right? $\endgroup$ – MathsMy Oct 28 '13 at 2:19
  • $\begingroup$ @Dosomemaths Re your first comment: that is a sequence of complex numbers. $\endgroup$ – Pedro Tamaroff Oct 28 '13 at 2:21
  • $\begingroup$ In Complex numbers, convergence $\iff$ Cauchy. But in other sets, isn't it Convergence $\rightarrow$ Cauchy but not the other way? $\endgroup$ – MathsMy Oct 28 '13 at 2:25
  • $\begingroup$ @Dosomemaths Sure. But you're assuming your function goes into $\Bbb C$. $\endgroup$ – Pedro Tamaroff Oct 28 '13 at 2:26

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