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Assume you have a linear functional $F:C^\alpha(\mathbb R^n) \mapsto\mathbb R$ such that $$ |F(f)| \leq \vert f \vert_{C^\alpha(\mathbb R^n)} $$ but only depending on the Holder seminorm, that is, $$ |f|_{C^\alpha} = \sup_{x,y}|x-y|^{-\alpha} |f(x)- f(y)|. $$

Can I assert that $F(f) = \int_{\mathbb R^n} g \cdot f \, dm$ for some $g \in H^p(\mathbb R^n)$ with $\Vert g \Vert_{H^p}\lesssim 1$ where $H^p$ is the real Hardy space? Is there any reference for Riesz representation theorems on Holder-Hardy spaces?

Edit: My idea why something like this might be true, is that we have $$ \int_{\mathbb R^n} f g \, dm \lesssim \vert f \vert_{C^\alpha} \Vert g \Vert_{H^p} $$ where $\Vert g \Vert_{H^p}$ is the atomic $H^p$ norm. Note that the dual of $H^p$ is the homogeneous space of Holder functions of order $n(1/p − 1)$.

HHN
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1 Answers1

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I don't think so. If you mod out constants, then the quotience space $C^\alpha(\mathbb{R})/\sim$ becomes a normed space with the seminorm $|\cdot|_{C^\alpha}$. Note that if $f\sim g$, that is, if $f-g$ is a constant, then $F(f)-F(g)=0$ by your inequality. Hence, if you define $L([f])=F(f)$ you have a well defined continuous linear functional in the dual of $C^\alpha(\mathbb{R})/\sim$. The dual of a normed space is a Banach space $Y$. When $p<1$ the real Hardy space $H^p(\mathbb{R}^n)$ is a topological vector space which is not locally convex (I read this in the book of Folland and Stein Hardy spaces on homogeneous groups, page 77, but there is no proof), this means in particular, that there is no norm on that space. Thus $Y$ cannot be $H^p$.

Concerning the dual of the space of Holder continuous functions, there is this interesting post, which indicates that the dual of the space of Holder continuous functions should be a space of measures.

Gio67
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  • I think it is possible to endow dual Banach space with topology which is not not locally convex. – dsh Sep 22 '23 at 22:08
  • You can change the topology in any space but in this case, you want to keep all the continuous linear functions continuous. I don't think this is possible. – Gio67 Sep 30 '23 at 11:46
  • If $\Vert g \Vert_{H^p}\lesssim 1$ condition is not imposed from the start for required functional, then, I think, it is still be possible that $H^p(\mathbb{R}^n)$ contains $(C^\alpha(\mathbb{R}))'$ as some Banachable subspace. For example the space of measurable functions, which is not generally locally convex, contains $L^p, p\ge 1,$ which are locally convex. Topology on space $(C^\alpha(\mathbb{R}))'$ of continuous linear functionals on space $C^\alpha(\mathbb{R})$ is a separate matter from them being continuous on $C^\alpha(\mathbb{R}).$ – dsh Sep 30 '23 at 15:59
  • Ok. I understand what you meant now. In this case, we have that $(H^p)’=C^\alpha$ and so $H^p\subset (H^p)’’=(C^\alpha)’$. So if $H^p$ contained the dual of $C^\alpha$, you would have equality. – Gio67 Oct 01 '23 at 10:02
  • If $(H^p)' = C^\alpha$ is not topological, then maybe $(H^p)''\neq (C^\alpha)'.$ I think that the proof cannot be easily extended to the case when $\Vert g \Vert_{H^p} \lesssim 1$ is not assumed. – dsh Oct 01 '23 at 11:24