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Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be defined by, for example, $f(x) = \frac{1}{2}x^2$. Then $f$ belongs to the Holder space $C^{1,1}(\mathbb{R})$.

Since $C^{1,1}(\mathbb{R})\subset C^{1,\alpha}(\mathbb{R})$ for $\alpha < 1$, then $f \in C^{1,\alpha}(\mathbb{R})$.

But this implies that $\frac{|f'(x) - f'(y)|}{|x-y|^{\alpha}} = |x-y|^{1-\alpha} \leq C$, for all $x, y$.

What is the obvious thing I am missing here? It seems the inclusion $C^{1,1}(\mathbb{R})\subset C^{1,\alpha}(\mathbb{R})$ is false, but I am looking at it right now on page 16 of the book `Introduction to the Calculus of Variations' by Bernard Dacorogna: http://www.amazon.com/Introduction-To-The-Calculus-Variations/dp/1860945082.

The precise statement in there is

Let $\Omega \subset \mathbb{R}^n$ be open. If $0 \leq \alpha \leq \beta \leq 1$ and $k \geq 0$ is an integer, then \begin{equation} C^{k,1}(\bar{\Omega}) \subset C^{k,\beta}(\bar{\Omega}) \subset C^{k,\alpha}(\bar{\Omega})\subset C^{k}(\bar{\Omega}). \end{equation}

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  • He is probably tacitly assuming (as also the notation $C^{k,\beta}(\overline{\Omega})$ suggests (note the "bar")), that $\Omega$ is bounded. In that case, your problem vanishes. – PhoemueX Jun 21 '14 at 05:13

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The inclusion is false on unbounded domains, unless one modifies the definition of the Hölder space, e.g., by replacing $$\frac{|u(x)-u(y)|}{|x-y|^\alpha}$$ with $$\frac{|u(x)-u(y)|}{|x-y|^\alpha+|x+y| }$$ (which would not be an unreasonable thing to do. Estimates of the kind $\lesssim |x-y|^\alpha$ are usually meant to be local; the consideration of large distances is mostly a nuisance.)

When reading Dacorogna's book, assume throughout that $\Omega$ is bounded. This came up before.

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