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Let $x \geq 0,c<0,$ and a Brownian motion $(W_u)_{u}.$ Let $T:=\inf\{u \geq 0, B_u +cu\geq x\}.$

It follows that $Y:=\sup_{u \geq 0}(B_u+cu) \in ]0,\infty[.$

We want to verify that $\{Y \geq x\} \subset (T<\infty).$

Supposing that $Y(w) > x$ then the result follows.

But what if $Y(w)=x,$ I can't see how to deduce it ? Do we need to use the continuity of the BM?

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Suppose $T=\infty$. Then $B_u+cu <x$ for all $u$ and $B_{u_n}+cu_n \to x$ for some sequenece $u_n \to \infty$ (because $Y=x$). But the $\frac {B_{u_n}} {u_n}+c \to 0$. This is a contradiction becasue $\frac {B_{u_n}} {u_n} \to 0$ (a.s.)and $c <0$.

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  • $\begingroup$ If $Y(w)=x,$ can we claim that in this case $Y(w) \in \{B_u+cu,u \geq 0\}$ ? $\endgroup$
    – Riro
    Commented May 14, 2022 at 0:15
  • $\begingroup$ No, you cannot. But there is a sequence in the set converging to $Y(\omega)$. $\endgroup$ Commented May 14, 2022 at 0:18
  • $\begingroup$ If there is a sequence $u_n$ such that $Y(w)$ is the limit of $B_{u_n}+cu_n,$ it possible to prove that $u_n$ is convergent (or has subsequence) ? (We are not supposing that $T(w)=\infty$) $\endgroup$
    – Riro
    Commented May 14, 2022 at 0:42
  • $\begingroup$ If $T(\omega) <\infty$ then $(u_n)$ can converge to a finite limit. But if $T(\omega)=\infty$ the $u_n$ must tend to $\infty$. @Riro $\endgroup$ Commented May 14, 2022 at 4:56

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