5 Appendix: Proof of Proposition 2.2

[126.1.1] The proof given below follows Ref. [30]. [126.1.2] Suppose $\lim _{{n\to\infty}}\mu _{n}(s)=\mu(s)$ and $\lim _{{n\to\infty}}\mu _{n}(a_{n}s+b_{n})=\nu(s)$ with $\mu(s)$ and $\nu(s)$ both nondegenerate. [126.1.3] Then it must be shown that there exist $a>0$ and $b$ such that

$\mu(s)=\nu(as+b).$

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[126.1.4] Pick a sequence of integers $n_{1}<n_{2}<\ldots<n_{k}<\ldots$ such that $\lim _{{k\to\infty}}a_{{n_{k}}}=a$ and $\lim _{{k\to\infty}}b_{{n_{k}}}=b$ exist with $0\leq a\leq\infty$ and $-\infty\leq b\leq\infty$ . [126.1.5] Consider this sequence of indices from now on as fixed. [126.1.6] Then, to simplify the notation, suppose without loss of generality that $\lim _{{k\to\infty}}a_{k}=a$ and $\lim _{{k\to\infty}}b_{k}=b$ .

[126.2.1] First it will be shown that $0<a<\infty$ . [126.2.2] Suppose $a=\infty$ . [126.2.3] Let

$u=\sup\{ x:\limsup _{{n\to\infty}}(a_{n}x+b_{n})<\infty\}.$

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[126.2.4] Then for $v<x<u$

$\limsup _{{n\to\infty}}(a_{n}v+b_{n})\leq\limsup _{{n\to\infty}}(v-x)a_{n}+\limsup _{{n\to\infty}}(a_{n}x+b_{n}),$

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and hence for every $v<u$ it follows that $\nu(v)=0$ because $(a_{n}v+b_{n})\to-\infty$ with $n\to\infty$ . [126.2.5] For $v>u$ , on the other hand, $\limsup(a_{n}v+b_{n})=\infty$ and hence $\nu(v)=1$ for $v>u$ . [126.2.6] Thus the assumption $a=\infty$ contradicts to $\nu(s)$ being nondegenerate.

[126.3.1] It follows that also $b$ must be finite. [126.3.2] In fact if $\lim _{{n\to\infty}}(a_{n}x+b_{n})=\infty$ then $\nu(x)=1$ while for $\lim _{{n\to\infty}}(a_{n}x+b_{n})=-\infty$ follows $\nu(x)=0$ .

[126.4.1] Suppose now that $a=0$ . [126.4.2] Then for every $x$ and $\varepsilon>0$

$b-\varepsilon\leq a_{n}x+b_{n}\leq b+\varepsilon$

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[page 127, §0] if $n$ is chosen sufficiently large. [127.0.1] By monotonicity of $\mu _{n}$ it follows that

$\mu _{n}(b-\varepsilon)\leq\mu _{n}(a_{n}x+b_{n})\leq\mu _{n}(b+\varepsilon).$

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[127.0.2] If $\varepsilon$ is chosen so that $\mu(x)$ is continuous at the points $b-\varepsilon$ and $b+\varepsilon$ , then

$\mu(b-\varepsilon)\leq\nu(x)\leq\mu(b+\varepsilon).$

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[127.0.3] Because $x$ was arbitrary it follows that $\mu(b-\varepsilon)=0$ and $\mu(b+\varepsilon)=1$ . [127.0.4] Hence $\mu(x)$ is degenerate, contrary to the conditions above.

[127.1.1] Finally, let $x$ be such that $\mu(x)$ is continuous at the point $ax+b$ , and that $\nu(x)$ is continuous at $x$ . [127.1.2] Then

$\lim _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})=\nu(x).$

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[127.1.3] On the other hand because $\lim _{{n\to\infty}}(a_{n}x+b_{n})=ax+b$ one has for sufficiently large $n$ that

$ax+b-\varepsilon\leq a_{n}x+b_{n}\leq ax+b+\varepsilon,$

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where $\varepsilon>0$ is chosen such that the distribution function $\mu$ is continuous at the points $ax+b-\varepsilon$ and $ax+b+\varepsilon$ . [127.1.4] Hence by monotonicity

$\mu _{n}(ax+b-\varepsilon)\leq\mu _{n}(a_{n}x+b_{n})\leq\mu _{n}(ax+b+\varepsilon)$

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and for $n\to\infty$

$\mu(ax+b-\varepsilon)\leq\liminf _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})\leq\limsup _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})\leq\mu(ax+b+\varepsilon).$

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[127.1.5] Because $ax+b$ is a point of continuity for $\mu(x)$ and $\varepsilon$ is arbitrary it follows that

$\lim _{{n\to\infty}}\mu _{n}(a_{n}x+b_{n})=\mu(ax+b)$

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and hence $\nu(x)=\mu(ax+b)$ proving the assertion.

Author:	R. Hilfer
originally published in:	Applications of Fractional Calculus in Physics, World Scientific, Singapore, page 87-130 (2000), ISBN 981-02-3457-0
originally published:	March, 2000