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I've read in a few papers that for $\lambda \in (0,1)$ the set of functions given by

$$c^{0,\lambda}([0,1]) = \lbrace f:[0,1]\to\mathbb{R}: \lim_{x\to y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}} = 0 \rbrace $$

is a closed subspace of

$$C^{0,\lambda}([0,1]) = \lbrace f:[0,1]\to\mathbb{R} : \sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}} < \infty \rbrace$$

equipped with the norm $\|f\|_{C^{0,\lambda}([0,1])} = \|f\|_{\infty}+\sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\lambda}}$.

However, I can not find a proof for this. Any help would be greatly appreciated for an explanation!

Richard Zhou
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  • First notice that you can get rid of the absolute value in the numerator of the limit in the first set. Then, you can see that the first set is the kernel of a linear functional, the limit. Use the definition of the norm in the second set to show that the linear functional is continuous. – Nathanson Mar 28 '15 at 01:07
  • @Nathanson No, the "little" space is much smaller than the kernel of any linear functional (it's a separable subspace of a nonseparable space). Your argument ignores the presence of $y$ in the definition. –  Mar 28 '15 at 02:36
  • @Woodface If $\lim_{x\to y}\frac{f(x)-f(y)}{|x-y|^{\lambda}}=\lim_{x\to y}\frac{g(x)-g(y)}{|x-y|^{\lambda}}=0$ then $0=a\lim_{x\to y}\frac{f(x)-f(y)}{|x-y|^{\lambda}}+b\lim_{x\to y}\frac{f(x)-f(y)}{|x-y|^{\lambda}}$$=\lim_{x\to y}\frac{(af+bg)(x)-(af+bg)(y)}{|x-y|^{\lambda}}$. – Nathanson Mar 28 '15 at 02:39

1 Answers1

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Introduce the modulus of continuity $$ \omega_f(\delta)= \sup_{|x-y|\le \delta}|f(x)-f(y)| $$ and observe that $$\omega_f(\delta)\le \|f\|_{C^{0,\lambda}}\delta^\lambda\tag{1}$$ $$f\in c^{0,\lambda} \iff \lim_{\delta\to0} \delta^{-\lambda}\omega_f(\delta)=0 \tag{2}$$

Given $f\notin c^{0,\lambda}$, let $r = \limsup_{\delta\to0} \delta^{-\lambda}\omega_f(\delta)>0$. If $g\in c^{0,\lambda}$, then by the triangle inequality $$ \limsup_{\delta\to0} \delta^{-\lambda}\omega_{f-g}(\delta) = \limsup_{\delta\to0} \delta^{-\lambda}\omega_{f}(\delta) = r $$ hence $\|f-g\|_{C^{0,\lambda}}\ge r$. This implies that the complement of $c^{0,\lambda}$ is open.