2
$\begingroup$

I am reading some literature where they refer to

"the Banach space of all $C^1$ maps from $[0,1]$ to $\mathbb{R}^n$."

But they don't say what norm they are using to make this into a Banach space. Is there a natural/standard choice of norm on this space which works?

My guess is that $$\|f\|=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|.$$ But I am not sure if this gives a Banach space. Is it? (Additionally, I would be interested to see a textbook reference of this material.)

$\endgroup$
4
$\begingroup$

Say you have a Cauchy sequence $\{f_n\}$ in the $C^1$ norm you propose. Then in particular $\{f_n\}$ is Cauchy in $C^0$ and so converges uniformly to a continuous function $f$. Likewise $\{f_n'\}$ is Cauchy in $C^0$ and converges uniformly to a continuous function $g$. Since $$f_n(x) = f_n(0) + \int_0^x f_n'(t) \, dt,\quad x \in [0,1]$$ holds for all $n$, you can let $n \to \infty$ and get $$f(x) = f(0) + \int_0^x g(t) \, dt,\quad x \in [0,1].$$ Thus $f \in C^1$ and $f' = g$ so that $f_n \to f$ in your $C^1$ norm.

$\endgroup$
  • $\begingroup$ this does not answer the op's question (which norm are you using). $\endgroup$ – Thomas Sep 20 '16 at 14:42
  • $\begingroup$ Great, thanks. But would you say that it is the most natural choice of norm which makes it Banach? Do you know any other? $\endgroup$ – Raoul Dugais Sep 20 '16 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.