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In solving problem sheets for my upcoming functional analysis $2$ exam I encountered the following statement:

Functions $u \in C^{0,1/2}(\mathbb{R}^n)$ satisfy $\|u\|_{C^0(\mathbb{R}^n)} < \infty$.

Here $C^{0,\alpha}(\mathbb{R}^n)$ denotes the space of Hölder continuous functions for $0 < \alpha \leq 1$, i.e. $$C^{0,\alpha} := \{f \in C^0(\mathbb{R}^n) : [f]_{C^{0,\alpha}} < \infty\}$$ where $$[f]_{C^{}0,\alpha} := \sup_{x,y \in \mathbb{R}^n,x\neq y} \frac{|f(x) - f(y)|}{|x - y|^\alpha}$$

What I tried: $$\|u\|_{C^0} = \sup_{x \in \mathbb{R}^n} |u(x)| \leq \sup_{x \in \mathbb{R}^n}|u(x) - u(y)| + |u(y)| \leq\sup_{x \in \mathbb{R}^n}|x - y|^{1/2} [u]_{C^{0,1/2}}+ |u(y)|$$ which does not help much

hmakholm left over Monica
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TheGeekGreek
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  • Isn't $u(x)=\sqrt {|x|}$ a counterexample? – zhw. Jul 20 '18 at 14:33
  • @zhw. Thanks! Yes, I also was very suspicious about this statement. – TheGeekGreek Jul 20 '18 at 14:42
  • It seems that by definition, you are saying $C^{0,\alpha}(\mathbb R^n) \subset C^0(\mathbb R^n).$ How are you defining $C^0(\mathbb R^n)$? Is this all continuous functions $f: \mathbb R^n\to\mathbb R$, or is it just the continuous functions $f : \mathbb R^n \to \mathbb R$ such that $|f|{C^0(\mathbb R^n)} = \sup{x \in \mathbb R^n} \lvert f(x) \rvert < \infty$? In the latter case, this claim is trivially true; in the former case it is false as zhw's example shows. – User8128 Jul 20 '18 at 14:42
  • @User8128 My notes are not very clear about this...but it is stated that if $\Omega \subseteq \mathbb{R}^n$ is bounded, then $$|f|{C^0} + [f]{C^{0,\alpha}}$$ is a norm on $C^{0,\alpha}(\Omega)$. Thus I suspect, that we are just dealing with continuous functions and not bounded ones, since otherwise this would also be a norm for $\Omega$ unbounded. – TheGeekGreek Jul 20 '18 at 14:46

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