The problem is as follows: I want to show that the normed space $C^2[0,1]$ with norm defined as $$\lVert f\rVert:=\max_{t\in[0,1]}\{\lvert f(t)\rvert+\lvert f''(t)\rvert\}$$ is a Banach space (and I have shown that this is indeed a norm).

In order to show that this space is a Banach space, I want to show that this normed space is complete; i.e. all Cauchy sequences converge. So I thought about taking sequences of functions that are Cauchy sequences. The problem is that I don't know if I can, in addition, assume that the Cauchy-sequence are $C^r$-stable; i.e. the distances between the $r$-th derivatives (w.r.t. this norm) are bounded for arbitrarily small values of the norm. I also don't know if I'm even thinking in the right direction since at first sight this question doesn't seem to be that challenging. I think I miss some important theory of converging function w.r.t. its $r$-th derivatives (although I'm familiar with $C^r$ stability as described above). Any useful words are appreciated, thanks in advance.

  • $\begingroup$ I believe the proof will be equivalent to the one the one given in the answer to this question. $\endgroup$
    – K.Power
    Feb 16, 2019 at 23:44

1 Answer 1


It is a standard exercise to see that $C^2([0,1])$ is a Banach space for the norm $$\|f\|' = \max_t (|f(t)| + |f'(t)| + |f''(t)|).$$ You can use the same approach as the question linked in the comments once you can work with this norm (just adding an extra derivative).

The problem you may have here is that you only have control on the function and its second derivative, but not its first derivative. It is clear that $\|f\| \leq \|f\|'$. I will show that $\|f\|' \leq c \|f\|$ so that the norms are equivalent and it is enough to show that $C^2([0,1])$ is complete for $\|\cdot \|'$.

To do this, note that for $y \geq 1/2$, we can taylor expand $$f(0) = f(y) - yf'(y) + \frac{y^2}{2} f''(\xi)$$ for some $\xi \in[\frac12,1)$. This implies that $$|f'(y)| \leq |y|^{-1} (|f(0)| + |f(y)|) + \frac{|y|}{2} |f''(\xi)| \leq 4 \|f\| + \frac{1}{2} \|f\|.$$ Similarly, for $y < \frac{1}{2}$, write $$f(1) = f(y) + (1-y)f'(y) + \frac{(1-y)^2}{2} f''(\xi)$$ and rearrange to again get $$|f'(y)| \leq 4 \|f\| + \frac{1}{2} \|f\|.$$ All in all, we've proved that $\|f'\|_\infty \leq c_1 \|f\|$. This means that $$\|f\|' = \max(|f(t)| + |f'(t)| + |f''(t)|) \leq \max(|f(t)| + \|f'\|_\infty + |f''(t)|) \leq \|f\| + c_1\|f\|$$ which is what we wanted to show.

  • $\begingroup$ I do not understand your $y$ and $\xi$ you introduce. $\endgroup$
    – Algebear
    Feb 17, 2019 at 20:27
  • $\begingroup$ $y$ is just a dummy variable in $[0,1]$. We want to show that $\|f'\|_\infty$ is bounded above by $c \|f\|$ for some $c$ so this means we want to bound $|f'(y)|$ for each $y \in [0,1]$. For $y \geq 1/2$, $\xi$ is the value arising in the remainder term when we taylor expand $f(0)$ around $y$ up to second order. Likewise, for $y< 1/2$, $\xi$ is the value obtained when we taylor expand $f(1)$ around $y$. $\endgroup$ Feb 17, 2019 at 20:41
  • $\begingroup$ Okay, I that's what I thought in the first place. However, I wondered why you didn't just stuck to $t$, but I guess that wouldn't matter. $\endgroup$
    – Algebear
    Feb 17, 2019 at 21:11
  • $\begingroup$ @RhysSteele true, but I don't like editing answers of other people ;) I delete my comment. Cheers! $\endgroup$
    – rae306
    Mar 29, 2020 at 19:58

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