Just a quick question on completeness.(adsbygoogle = window.adsbygoogle || []).push({});

The space of all continuous complex values functions [itex]\mathcal{C}([0,1])[/itex] with the norm

[tex]\|f\| = \sup\{|f(x)| \,:\, x \in [0,1]\}[/tex]

is a Banach space because every Cauchy sequence in [itex]\mathcal{C}([0,1])[/itex] converges to a limit point in [itex]\mathcal{C}([0,1])[/itex].

My question is this: What if the norm on [itex]\mathcal{C}([0,1])[/itex] is instead defined as

[tex]\|f\| = \int_0^1 |f(x)|dx[/tex]

This is a perfectly good norm (as it satisfies the 3 axioms), but does it cause [itex]\mathcal{C}([0,1])[/itex] to be complete?

My guess is no. But Im not sure if my solution is very good. Here it is.

Take a Cauchy sequence [itex]\{f_n\}[/itex] in [itex]\mathcal{C}([0,1])[/itex] and write

[tex]f_n = \{f_1,f_2,\dots\}[/tex].

So [itex]\{f_n\}[/itex] is just a sequence of continuous complex valued functions. I want to show that this converges in [itex]\mathcal{C}([0,1])[/itex], that is, the distance (as defined by the norm) between successive terms approaches zero the further along the sequence you go. More precisely:

Given any [itex]\epsilon > 0[/itex], we need to find an [itex]N[/itex] such that for all [itex]m,n > N[/itex]

[tex]\|f_n - f_m\| < \epsilon[/tex]

This implies

[tex]\int_0^1 |f_n(x) - f_m(x)|dx < \epsilon[/tex]

[tex]\int_0^1 |f_n(x)|dx - \int_0^1|f_m(x)|dx < \epsilon[/tex]

[tex]\|f_n\| - \|f_m\| < \epsilon[/tex]

This is where I am struggling. I have the notion that

[tex]\|f_n\| - \|f_m\| \rightarrow 0[/tex]

Does not necessarily mean that [itex]\|f_n\| \rightarrow 0[/itex]. In other words, I dont see how this norm makes [itex]\mathcal{C}([0,1])[/itex] a Banach space because the distance between successiver terms may approach zero regardless of whether [itex]m,n \rightarrow \infty[/itex].

This is a fairly straight-forward exercise in completeness, but I am struggling to come to a clear and concise proof. Any help is appreciated.

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# Homework Help: Just a quick question on completeness

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