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For $0<\alpha\leq 1$ let $\Lambda_{\alpha}([0,1])$ be the space of functions on $[0,1]$ such that $||{f_{\Lambda_{\alpha}}}||<\infty$, where $$\|{f}\|_{\Lambda_{\alpha}}=|f(0)|+\text{sup}_{x,y\in[0,1],x\neq y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$

For $0<\alpha<\beta\leq 1$ the Banach space $\Lambda_{\beta}([0,1])$ is a proper subspace of $\Lambda_{\alpha}([0,1])$ (consider $f(x)=x^\alpha$), but I'm still trying to prove that it is not dense in $\Lambda_{\alpha}([0,1])$. Any suggestions would be greatly appreciated, thanks!

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The same example $f(x)=x^\alpha$ helps again. If $g\in \Lambda^\beta$, then let $h=f-g$ and observe that $$ \frac{|h(x)-h(0)|}{|x-0|^\alpha}\ge \frac{|f(x)-f(0)|}{|x|^\alpha} - \frac{|g(x)-g(0)|}{|x|^\alpha} = 1 - \frac{|g(x)-g(0)|}{|x|^\alpha} $$ where the quotient $\frac{|g(x)-g(0)|}{|x|^\alpha} $ tends to $0$ as $x\to 0$. Therefore, $\|f-g\|_{\Lambda^\alpha}\ge 1$.

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  • $\begingroup$ Great, thanks a lot! $\endgroup$ – Aubrey Feb 1 '15 at 16:54

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