0
$\begingroup$

$C[a,b]$ 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).

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ "$\langle f_n(x)\rangle$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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