2

I have a normed space (I'll denote it $C^2[a;b]$) which consists of continuous real functions whose first and second derivatives are also continuous in interval $[a;b]$.

$\forall x,y \in C^2[a;b]$ and $\forall \lambda \in \mathbb{R}$ the norm is defined as follows

$$ \|x(t)\| = |x(a)| + \max_{t \in [a;b]} | x'(t) | + \sqrt{\int_a^b (x''(t))^2dt} ;$$

It is in fact a valid norm following the definition.

But do I have a Banach space? It should follow from definition as well but I have little experience with such formal arguments.

Pranasas
  • 1,370
  • Why did you close your new question on the same topic? If you have a solution, just post it as an answer, I'd really like to see it. – savick01 Aug 06 '13 at 20:38
  • @savick01 Regarding my new question I was curious whether one could similarly exploit the Banach space situation when the first derivative is omitted in the definition of my norm. However, as it was quickly pointed out, without the first derivative it is not even a norm (only a seminorm) so the idea failed. – Pranasas Aug 06 '13 at 20:52
  • Are you sure? I think that something like $|x(a)|+|x(b)|+\max x''$ would be a norm. (It is enough to take two different values of $x$, $x'$ is unnecessary.) – savick01 Aug 06 '13 at 20:57
  • I think it would also be a Banach space since the norm is roughly just the supremum norm applied to the second derivative. – savick01 Aug 06 '13 at 21:01

1 Answers1

1

It's not. Start with $z_n(t)$ being continuous, zero for $-1\le t\le0$, one for $1/n\le t\le1$, and linear on $[0,1/n]$. Put $y_n(t)=\int_{-1}^t z_n(\tau)\,d\tau$ and $x_n(t)=\int_{-1}^t y_n(\tau)\,d\tau$.

You now have a Cauchy sequence in your space (on the interval $[-1,1]$ which does not converge. Its limit ought to have been zero for $t<0$ and $\frac12t^2$ for $t\ge0$, and that function is not in your space, since the second derivative is not continuous.

Harald Hanche-Olsen
  • 31,960
  • But where does the specific norm come into play? Looking at your proof it seems that there will be no Banach space no matter what the norm is... – Pranasas May 01 '13 at 17:25
  • That's because I skipped some detail. If you had used a maximum norm on the second derivative, then this would not be an example, since $x_n''=z_n$ does not converge uniformly. It does converge in $L^2$ however, and that is what counts in the present case. – Harald Hanche-Olsen May 01 '13 at 18:20
  • So, roughly speaking, it's precisely the integral in my norm definition which is responsible for allowing this counter-example, correct? – Pranasas May 01 '13 at 18:41
  • Indeed it is so. – Harald Hanche-Olsen May 01 '13 at 18:51