Let $S$ be the unit sphere of $C[0,1], \|\cdot\|_{\infty})$, let $(B_{t})_{t}$ the brownian motion.
I would like to show that the measure $\mu_r$ defined on $\mathbb{B}(S)$ by
$\mu_r(A):=P\Big(\frac{B}{\| B \|_{\infty}}\in A \Big| \|B\|_{\infty}>r\Big)$ does not converge weakly to any measure as $r \to \infty$.
I tried to evaluate the measure $\mu$ on the subsets of the cylinder generated by the evaluation map but it is still difficult to manipulate.
Does any one have an idea ?
These measures do converge weakly to a measure with two atoms (of equal probabilities $1/2$) at the two paths $\phi_{\pm}:t\mapsto \pm t$. One way to see it is to consider $\frac{1}{r}B_t$, so that the condition is $||\cdot||_\infty\geq 1$, and apply Schilder's theorem: observe that $\phi_{\pm}$ are the only minimizers of the rate function $I(\phi)=\int_0^1 \phi'(t)^2\,dt$ in the set $S_0:=\{\phi:||\phi||_\infty\geq 1,\,\phi(0)=0\}$. On the other hand, for every positive $\epsilon>0$, we have $I((1+\epsilon)\phi_+)=(1+\epsilon)^2$, and an open $\epsilon$-neighborhood of $(1+\epsilon)\phi_+$ is contained in $S_0$, hence
$$\liminf\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in S_0\right)\geq -1.$$
Since this rate function is good, in particular, lower semi-continuous under $||\cdot||_\infty$ norm, we have that the sequence of conditioned measures is tight: for every $\epsilon>0$, the set $E_{1+\epsilon}:=\{\phi:I(\phi)\leq 1+\epsilon\}$ is compact, and $$ \limsup\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}|\frac{1}{r}B\in S_0\right)\leq \limsup\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}\right)-\liminf\frac{1}{r^2}\mathbb{P}\left(\frac{1}{r}B\in S_0\right)\leq -\inf_{\phi\in \overline{E_{1+\epsilon}}}I(\phi)+1= -\epsilon, $$ in particular, $\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}|\frac{1}{r}B\in S_0\right)\to 0$, proving tightness (at the end, we used lower continuity again).
On the other hand, for any closed set $F\subset S_0\setminus\{\phi_{\pm}\}$, we have, by lower semi-continuity again, $F\subset E^c_{1+\epsilon}$ for $\epsilon$ small enough. Therefore, portmanteau theorem implies that any subsequential limit gives probability $0$ to any such $F$, which by regularity of measures is enough to conclude that any subsequential limit is supported on $\phi_{\pm}$.
$\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\ep}{\varepsilon}$Here is an elementary proof that $\mu_r$ converges weakly (as $r\to\infty$) the measure $\mu$ that is the distribution of the stochastic process $Y$ given by the formula \begin{equation*} Y_t:=t\sgn B_1 \tag{10}\label{10} \end{equation*} for $t\in[0,1]$. Let also \begin{equation*} X:=\frac B{\|B\|}, \tag{20}\label{20} \end{equation*} where $\|\cdot\|:=\|\cdot\|_\infty$, and let $M:=\max_{t\in[0,1]}B_t$.
Then, by symmetry and the reflection principle,
\begin{equation*}
P(|B_1|\le r-1,\|B\|>r)\le2P(M>r,B_1\le r-1)\\
=2P(B_1>r+1)=o(P(B_1>r)) \tag{30}\label{30}
\end{equation*}
and
\begin{equation*}
P(\|B\|>r+1)\le2P(M>r+1)\\
=4P(B_1>r+1)=o(P(B_1>r)). \tag{40}\label{40}
\end{equation*}
On the other hand,
\begin{equation*}
P(\|B\|>r)\ge P(B_1>r). \tag{50}\label{50}
\end{equation*}
Let
$$D:=\|B\|-|B_1|,$$
so that $D\ge0$ and
\begin{equation*}
D>2\implies
\|B\|>r+1\text{ or }|B_1|<r-1. \tag{60}\label{60}
\end{equation*}
By \eqref{60}, \eqref{40}, \eqref{30}, and \eqref{50},
\begin{equation*}
P(D>2,\|B\|>r) \\
\le P(\|B\|>r+1)+P(|B_1|\le r-1,\|B\|>r)
=o(P(\|B\|>r)). \tag{70}\label{70}
\end{equation*}
Take now any real $\ep>0$. For $t\in[0,1]$, let \begin{equation*} B^0_t:=B_t-tB_1, \end{equation*} so that $B^0$ is a Brownian bridge independent of $B_1$. Let $D:=\|B\|-|B_1|$, so that $D\ge0$ and \begin{equation*} \begin{aligned} X_t-Y_t&=\frac{B_t}{|B_1|+D}-t\sgn B_1 \\ &=\frac{tB_1}{|B_1|+D}-t\sgn B_1+\frac{B^0_t}{|B_1|+D} \\ &=-t\sgn B_1\frac{D}{\|B\|}+\frac{B^0_t}{|B_1|+D}, \end{aligned} \end{equation*} so that \begin{equation*} \|X-Y\|\le \frac{D}{\|B\|}+\frac{\|B^0\|}{|B_1|}. \tag{80}\label{80} \end{equation*}
Next,
\begin{equation*}
P(\|X-Y\|>\ep,\|B\|>r)\le p_1+p_2,
\end{equation*}
where
\begin{equation*}
p_1:=P(D>2,\|B\|>r)=o(P(\|B\|>r))
\end{equation*}
by \eqref{70}, and, by \eqref{80}, for $r>\max(2,4/\ep)$
\begin{equation*}
\begin{aligned}
p_2&:=P(\|X-Y\|>\ep,D\le2,\|B\|>r) \\
&=P(\|X-Y\|>\ep,D\le2,\|B\|>r,|B_1|>r-2) \\
&\le P\Big(\frac2{\|B\|}>\frac\ep2,\|B\|>r\Big)
+P\Big(\frac{\|B^0\|}{r-2}>\frac\ep2,|B_1|>r-2\Big) \\
&=0+P\Big(\|B^0\|>\frac\ep2\,(r-2)\Big) P(|B_1|>r-2) \\
&=o(P(|B_1|>r))=o(P(\|B\|>r)).
\end{aligned}
\end{equation*}
So,
\begin{equation}
P\big(\|X-Y\|>\ep\big|\|B\|>r\big)=o(1). \tag{90}\label{90}
\end{equation}
Also, by symmetry, $Y$ is independent of $\|B\|$, so that the conditional distribution of $Y$ given $\|B\|>r$ is the same as the (unconditional) distribution of $Y$ (in $C[0,1]$).
It now follows from \eqref{90} that, as $r\to\infty$, the conditional distribution of $X$ given $\|B\|>r$ converges to the distribution of $Y$ in the Lévy–Prokhorov metric and hence weakly. $\quad\Box$
