4
$\begingroup$

We are on the line $[0,T]$, let $a \in (0,1)$. Let $C^a$ consist of those real functions such that:

$|f|_a=\sup_{0\le s

Denote the norm $\|f\|_a=|f|_a+|f|_\infty$. Where $|\cdot|_\infty$ is the standard supremum norm for continuous funtions.

Using the mean-value theorem, we can show that the space $\mathcal{C}^1$, the space of continuously differentiable functions, is in $C^a$.

A theorem in a book I am reading states:

Let $f \in C^a$ with $a \in (0,1)$, and let $b\in(0,1)$ be such that $a+b>1$. The linear operator $T_f: \mathcal{C}^1\subset C^b\rightarrow C^b$ defined as $Tf(g)=\int_0^\cdot f(u)g'(u)du$ is continuous with respect to the norm $\|\cdot\|_b$. By density, it extends (in a unique way) to an operator $T_f:C^b\rightarrow C^b.$

The author proves everything except from the density argument. From what I understand he refers to the classical result that if you have a dense subspace of a banach space, and you have a continous linear operator, you can extend it uniquely to the entire space? The problem I have is how is it dense. Because according to these links:

$C^1$ is not dense in Hölder space

Polynomials not dense in holder spaces

$\mathcal{C}^1$ is not dense in the Hölder space. I think in the links they only use the norm $|\cdot|_a$, not $\|\cdot\|_a$, but if it is not dense w.r.t to the first norm, it can't be with respect to the other norm that is a sum of the first norm and the supremum norm?

Do you see what I am missing? What kind of density argument is used here?

  • 0
    I guess it's just a mistake. $\mathcal{C}^1$ is dense in $\mathcal{C}^0$, so if you're not paying attention, it's easy to believe it would also be dense in $C^b$, because $C^b \subset \mathcal{C}^0$. Of course we have a unique continuous extension to the closure of $\mathcal{C}^1$, and by Hahn-Banach continuous extensions to $C^b$. But since $\mathcal{C}^1$ isn't dense in $C^b$, these continuous extensions aren't unique. It could be, however, that there's a unique extension with the same norm. – 2017-01-09
  • 0
    @DanielFischer Thank you but I quite did not understand that. How do you use the Hahn-Banach theorem to show this? I see that it talks about functionals, https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem , but here the value of the linear operator is a function not a real or complex number? And, would it be difficult to show the uniqueness you talk about, that there is a unique norm? – 2017-01-09
  • 0
    Brainfart on my part. Sorry. – 2017-01-09
  • 0
    @DanielFischer I think you are correct so no need to say sorry, I found this: https://en.wikipedia.org/wiki/Continuous_linear_extension and it talks about an extension the same way you do, and there Y may not be the real or complex numbers. So there is probably a way to do it. – 2017-01-09
  • 0
    We get the extension to the closure. That part is unproblematic. The question is whether we can extend to the whole space. If the codomain is finite-dimensional, Hahn-Banach gives the extension, but for infinite-dimensional codomain, an extension to the whole space might not exist. // $\mathcal{C}^1$ is complete in the $C^1$-norm, whether it's complete as a subspace of $C^b$ is a different question. – 2017-01-09
  • 0
    Can you give a reference to the book, in case we're missing some context? There are no useful norm-dense subsets of functions in $C^b$. The closure of $C^1$ in the $C^b$ norm is the little-oh space Hölder space $c^b$, which is a proper subspace of $C^b$ (of uncountable codimension). – 2017-01-09
  • 0
    @zaq http://www.springer.com/us/book/9788847028227, page 25. – 2017-01-09

1 Answers 1

2

The appeal to "density" is a little sloppy. I believe the following is what the author meant:

  1. The closure of $\mathcal C^1$ with respect to $C^\beta$ norm is the little Hölder space $c^{\beta}$ (defined by the condition $|f(x)-f(y)| = o(|x-y|^\beta)$ as $x-y\to 0$)
  2. $C^{\beta+\epsilon}\subset c^\beta$ for every $\epsilon>0$.
  3. Given $f\in C^\alpha$ and $\beta$ such that $\alpha+\beta>1$, pick $\gamma <\alpha$ such that $\gamma+\beta >1$. Then $f\in c^\gamma$, hence the density argument applies and shows that the integral $\int fg'$ is bounded by $C\|f\|_\gamma\|g\|_\beta$. The latter is controlled by $C\|f\|_\alpha\|g\|_\beta$, as desired.

I don't have a suitable reference for (1), but this fact is not hard to prove directly: (a) smooth functions are in $c^\beta$; (b) $c^\beta$ is a closed subspace of $C^\beta$; (c) mollifying a $c^\beta$ function produces an approximating sequence that converges in $C^\beta$ norm; the key is that sufficiently small scales don't really contribute to the norm.

For reference, here is the relevant part of the source (page 25 in Selected Aspects of Fractional Brownian Motion by Ivan Nourdin).

thm31

The reference Young [67] is to the classical article An inequality of the Hölder type, connected with Stieltjes integration, Acta Math. (1936) 67, 251-282. Young actually works with larger spaces of functions of finite $p$-variation, and I couldn't locate a relevant density argument there. A more accessible source is a pair of blog posts by Fabrice Baudoin:

  • 0
    Thank you very much for doing so much work and going and reading all the sources! I do not know so much about these spaces you mention, so I will try to learn more about it. – 2017-01-10