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Currently I'm trying to read Stroock's "lectures on topics in stochastic differential equations" book. In lemma $2.5$, he states that given $f$ and $g$ that are strictly increasing function on $[0, infty]$ such that $f(0)=g(0)=0$ and $g(infty)=infty$, for a normed linear space $L$ and function $h: mathbb{R}^d to L$ that is strongly cont. on open ball $B_{r}(a)$, if $$int_{B_{r}(a)}int_{B_{r}(a)}g(frac{||h(x)-h(y)||}{f(|x-y|)})dxdy leq B$$ then $$||h(x)-h(y)||leq 8 int_{0}^{|x-y|}g^{-1}(frac{4^{d+2}B}{alpha ^2 u^{2d}})P(du)$$, $$x,y in B_{r}(a)$$ where $$alpha = inf _{x in B_{r}(a)}inf_{1< m leq 2} frac{|B_{m}(x)cap B_{r}(a)|}{m^d} $$(I think by absolute value here we mean its volume).

He starts the proof by defining $$I(x)= int_{B_{r}(a)} g(frac{||h(x)-h(y)||}{f(|x-y|)})dy$$, and then picking distinct $x,y in B_{r}(a)$ and $n=|x-y|$. Then says we can choose $c in B_{n/2}((x+y)/2)$ such that

$$I(c) leq frac{2^{d+1}B}{alpha n^d}$$ because $$int_{B_{r}(a)}I(c)dc leq B$$

My problem is that I couldn't deduce the bound given for $I(c)$ here or why we can choose such $c$. I think he uses the last inequality to get the bound but then it would mean that $$Bint_{B_{r}(a)} 1 dx leq frac{2^{d+1}B}{alpha n^d}$$, but this doesn't makes sense for me either. I appreciate any help.

real-analysis functional-analysis

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edited Mar 20 at 18:46

dankmemer

asked Mar 20 at 18:22

dankmemerdankmemer

149113

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Currently I'm trying to read Stroock's "lectures on topics in stochastic differential equations" book. In lemma $2.5$, he states that given $f$ and $g$ that are strictly increasing function on $[0, infty]$ such that $f(0)=g(0)=0$ and $g(infty)=infty$, for a normed linear space $L$ and function $h: mathbb{R}^d to L$ that is strongly cont. on open ball $B_{r}(a)$, if $$int_{B_{r}(a)}int_{B_{r}(a)}g(frac{||h(x)-h(y)||}{f(|x-y|)})dxdy leq B$$ then $$||h(x)-h(y)||leq 8 int_{0}^{|x-y|}g^{-1}(frac{4^{d+2}B}{alpha ^2 u^{2d}})P(du)$$, $$x,y in B_{r}(a)$$ where $$alpha = inf _{x in B_{r}(a)}inf_{1< m leq 2} frac{|B_{m}(x)cap B_{r}(a)|}{m^d} $$(I think by absolute value here we mean its volume).

He starts the proof by defining $$I(x)= int_{B_{r}(a)} g(frac{||h(x)-h(y)||}{f(|x-y|)})dy$$, and then picking distinct $x,y in B_{r}(a)$ and $n=|x-y|$. Then says we can choose $c in B_{n/2}((x+y)/2)$ such that

$$I(c) leq frac{2^{d+1}B}{alpha n^d}$$ because $$int_{B_{r}(a)}I(c)dc leq B$$

My problem is that I couldn't deduce the bound given for $I(c)$ here or why we can choose such $c$. I think he uses the last inequality to get the bound but then it would mean that $$Bint_{B_{r}(a)} 1 dx leq frac{2^{d+1}B}{alpha n^d}$$, but this doesn't makes sense for me either. I appreciate any help.

real-analysis functional-analysis

share|cite|improve this question

edited Mar 20 at 18:46

dankmemer

asked Mar 20 at 18:22

dankmemerdankmemer

149113

$endgroup$

add a comment | 

$begingroup$

Currently I'm trying to read Stroock's "lectures on topics in stochastic differential equations" book. In lemma $2.5$, he states that given $f$ and $g$ that are strictly increasing function on $[0, infty]$ such that $f(0)=g(0)=0$ and $g(infty)=infty$, for a normed linear space $L$ and function $h: mathbb{R}^d to L$ that is strongly cont. on open ball $B_{r}(a)$, if $$int_{B_{r}(a)}int_{B_{r}(a)}g(frac{||h(x)-h(y)||}{f(|x-y|)})dxdy leq B$$ then $$||h(x)-h(y)||leq 8 int_{0}^{|x-y|}g^{-1}(frac{4^{d+2}B}{alpha ^2 u^{2d}})P(du)$$, $$x,y in B_{r}(a)$$ where $$alpha = inf _{x in B_{r}(a)}inf_{1< m leq 2} frac{|B_{m}(x)cap B_{r}(a)|}{m^d} $$(I think by absolute value here we mean its volume).

He starts the proof by defining $$I(x)= int_{B_{r}(a)} g(frac{||h(x)-h(y)||}{f(|x-y|)})dy$$, and then picking distinct $x,y in B_{r}(a)$ and $n=|x-y|$. Then says we can choose $c in B_{n/2}((x+y)/2)$ such that

$$I(c) leq frac{2^{d+1}B}{alpha n^d}$$ because $$int_{B_{r}(a)}I(c)dc leq B$$

My problem is that I couldn't deduce the bound given for $I(c)$ here or why we can choose such $c$. I think he uses the last inequality to get the bound but then it would mean that $$Bint_{B_{r}(a)} 1 dx leq frac{2^{d+1}B}{alpha n^d}$$, but this doesn't makes sense for me either. I appreciate any help.

real-analysis functional-analysis

share|cite|improve this question

edited Mar 20 at 18:46

dankmemer

asked Mar 20 at 18:22

dankmemerdankmemer

149113

$endgroup$

share|cite|improve this question

edited Mar 20 at 18:46

dankmemer

asked Mar 20 at 18:22

dankmemerdankmemer

149113

share|cite|improve this question

edited Mar 20 at 18:46

dankmemer

asked Mar 20 at 18:22

dankmemerdankmemer

149113

asked Mar 20 at 18:22

dankmemerdankmemer

149113

asked Mar 20 at 18:22

dankmemerdankmemer

149113