# Showing a space is not Banach

• Mar 28th 2010, 01:45 PM
vinnie100
Showing a space is not Banach
Let C[0,1] be the space of continuous real valued functions on [0.1] which has a continuous derviative i.e. its C1.

Define the norm ||f|| = |f(0)| + sup |f '(t)| for t in [0,1]

How do you show that this is not a Banach space?

I think I might I have a counterexample though I cannot prove by defn of Cauchy sequences that it does not converge.

So we would want to find a Cauchy sequence that does not converge.
i.e. ||f_n - f_m|| = |f_n(0) - f_m(0)| + sup |f '_n(0) - f '_m(0)|

I was thinking f_n(t) = t^n would work but then I could not figure it out.
• Mar 28th 2010, 03:23 PM
Jose27
Quote:

Originally Posted by vinnie100
Let C[0,1] be the space of continuous real valued functions on [0.1] which has a continuous derviative i.e. its C1.

Define the norm ||f|| = |f(0)| + sup |f '(t)| for t in [0,1]

How do you show that this is not a Banach space?

I think I might I have a counterexample though I cannot prove by defn of Cauchy sequences that it does not converge.

So we would want to find a Cauchy sequence that does not converge.
i.e. ||f_n - f_m|| = |f_n(0) - f_m(0)| + sup |f '_n(0) - f '_m(0)|

I was thinking f_n(t) = t^n would work but then I could not figure it out.

Maybe I'm missing something but I think this is a Banach space:

Let $(u_n)$ be a Cauchy sequence then $(u_n(0))$ is a Cauchy sequence and thus converges and $(u'_n)$ is a unif. Cauchy sequence so it also converges, and by a standard theorem we have that $u$ is a $C^1$ function and $u'_n \rightarrow u'$ if $u_n\rightarrow u$