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I have to show that the $C^1[0,1]$ is not complete with any of these norms:

  • $\|f\|_{\infty}=\sup_{x\in[0,1]}|f(x)|$
  • $\|f\|_{*}=|f(0)|+\int_0^1|f'(x)|dx$

My attempt

The right sequence for the first norm is $f_n=\sqrt{x+\frac{1}{n}}$.

Notice that $\forall n\in\mathbb{N} : f_n\in C^1[0,1]$

Let $f=\sqrt{x}$

We see that $(f_n)$ converges to $f$ in sup norm in $C[0,1]$, thus it is Cauchy.

$C^1[0,1]$ is a subspace of $C[0,1]$ and all terms of $(f_n)$ are in $C^1[0,1]$, so $(f_n)$ is Cauchy in $C^1[0,1]$

But $f$ is not in $C^1[0,1]$. So $C^1[0,1]$ with sup norm is not complete.

When it comes to the second norm, I think the same sequence will be also okay. Am I right?

luka5z
  • 6,359

1 Answers1

5

Yes. Your same functions should work for $\| \cdot \|_*$. To be sure, note that if $n > m$, then $|f'_n(x) - f'_m(x)| = f'_n(x) - f'_m(x)$. Therefore, $$\int_0^1 |f'_m(x) - f'_n(x)| \,dx = \int_0^1 f'_n(x)\,dx - \int_0^1 f'_m(x) \, dx = f_n(1)-f_m(1) + f_m(0) - f_n(0) \\ = \sqrt{1+\frac{1}{n}} - \sqrt{1+\frac{1}{m}} + \sqrt{\frac{1}{m}} - \sqrt{\frac{1}{n}}.$$

From this you can deduce that for any $n, m \in \Bbb{N}$ $$\| f_n(x) - f_m(x) \|_* \leq 2\left| \sqrt{\frac{1}{n}} - \sqrt{\frac{1}{m}} \right| + \left| \sqrt{1 + \frac{1}{n}} - \sqrt{1 + \frac{1}{m}} \right|$$ which goes to $0$ as $n, m \to \infty$, hence $\{f_n\}$ is Cauchy in $\| \cdot \|_*$.

Tom
  • 9,978