Riesz representation theorem for functionals acting on Hölder $C^\alpha$ functions

duality-theoremsfunctional-analysishardy-spacesharmonic-analysis

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 $n(1/p − 1)$.

Best Answer

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.