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Let $0<\alpha<\beta\le1$ and $-\infty<a<b<\infty$ then $C^{0,\beta}([a,b])\subsetneq C^{0,\alpha}([a,b])$

Proof:

First define $\|f\|_{C^{0,\alpha}}:=\|f\|_{C^0([a,b])}+[f]_{C^{0,\alpha}([a,b])}$ where $[f]_{C^{0,\alpha}([a,b])}:=\sup\limits_{a\le x\ne y\le b}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}$

$\sup\limits_{a\le x\ne y\le b}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\le^{\text{why not equality?}} \sup\limits_{a\le x\ne y\le b}\frac{|f(x)-f(y)|\cdot |x-y|^{\beta - \alpha}}{|x-y|^{\beta}}\le [f]_{C^{0,\beta}([a,b])}(b-a)^{\beta-\alpha}$

$\|f\|_{C^{0,\alpha}}\le \|f\|_{C^0}+[f]_{C^{0,\beta}}(b-a)^{\beta-\alpha}\le (1+(b-a)^{\beta-\alpha})\|f\|_{C^{0,\beta}}$

How to prove it is a strict inclusion? I am having a hard time visualizing what kind of function satisfies the Holder condition.

kimchi lover
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John Cataldo
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1 Answers1

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Assume that $a=0$ for simplicity and that $0<\alpha<\beta\le 1$. Consider the function $f(x)=x^\alpha$. Then taking $x=0$ and $y=\frac1n$ you have that $$\frac{|f(y)-f(x)|}{|y-x|^\beta}=\frac{\frac1{n^\alpha}}{\frac1{n^\beta}}=n^{\beta-\alpha}\to\infty,$$ which shows that $f$ is not in $C^{0,\beta}$. On the other hand, if $0\le x<y\le b,$ then $$\frac{|f(y)-f(x)|}{|y-x|^\alpha}=\frac{y^\alpha-x^\alpha}{(y-x)^\alpha}\le 1$$ since $f$ is subadditive (see for example here subadditivity)

Gio67
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