Let $F\subseteq C^1([0,1],\mathbb{R})$ be a closed subspace of $C([0,1],\mathbb{R})$. Show that $F$ is of finite dimension.

So, I considered the norm $\Vert f\Vert_1=\Vert f\Vert_\infty+\Vert f'\Vert_\infty$ and showed that $(F,\Vert\cdot\Vert_1)$ is a Banach space. Next, using Arzela-Ascoli, I showed that $\overline{\mathbb{B}}_{(F,\Vert\cdot\Vert_1)}(0,1)$ is compact in $(F,\Vert\cdot\Vert_\infty)$. At this point, I got help and was told to show that $\Vert\cdot\Vert_\infty$ and $\Vert\cdot\Vert_1$ are equivalent norms on $F$ and that I could use the open mapping theorem to do so... How?

The next step would be to use the Riesz compactness theorem to conclude that $F$ is, in fact, of finite dimension. But, in which step, exactly, did we need the closedness of $F$? I have the feeling of not having talked about it explicitly.


You have $\|x\|_{\infty}\leq \|x\|_1$, this implies that $Id:(C^1([0,1]),,\|\|_1)\rightarrow (C^1([0,1]),\|\|_{\infty}$ is continuous (a sequence which converges for $\|\|_1$ converges for $\|\|_{\infty}$), since it is surjective, it is an homemorphism (open mapping), and $Id^{-1}(B_{\|\|_{\infty}}(0,1))=B_{\|\|_{\infty}}(0,1)$ is open thus there exists $c>0$ such that $B_{\|\|_{1}}(0,c)\subset B_{\|\|_{\infty})}(0,1)$ and the norms are equivalent.

Hint: since $\|\|_{\infty}$ is equivalent to $\|\|_{1}$, $Id:C^1([0,1])\rightarrow C^1([0,1])$ is an homeomorphism, you deduce that $B_{F,\|\|_1}(0,1)$ is compact and $F$ is finite dimensional. You have to review the proof that $(F,\|\|_1)$ is Banach to see if you used the fact that $F$ is closed.

  • $\begingroup$ Thank you! But, how do I show the equivalence of the norms? Yes, you are right, but the question remains in the following form: where do I need that $F$ is Banach space and not just a normed vector space? It seems like a stupid question, but as we use a lot of theorems I just want to know about all the interconnections and uses of premises... $\endgroup$ – Ramen Jan 11 '18 at 0:27
  • 2
    $\begingroup$ @Ramen Check that the identity map is continuous from $F$ with the $\infty$-norm to $F$ with the $1$-norm. Then by the open mapping theorem, the identity map is an open mapping so in fact the identity map in the other direction is continuous and hence the norms are equivalent. To invoke the open mapping theorem you have used that $F$ is a Banach space and hence in particular that $F$ is closed. $\endgroup$ – Rhys Steele Jan 11 '18 at 0:36
  • $\begingroup$ @nobody Thank you. I would accept your comment as an answer. And, I'd perhaps rather show that the identity from $F$ with the $1$-norm to $F$ with the $\infty$-norm is continuous, no? $\endgroup$ – Ramen Jan 11 '18 at 0:51
  • $\begingroup$ @Ramen Yes, I meant for the norms to be the other way around. $\endgroup$ – Rhys Steele Jan 11 '18 at 1:02

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.