Brownian motion and Holder-$\frac{1}{2}$-continuity

Question

Let B be a Brownian motion. For every $K>0$, we have

$$
P[\inf \left \{ t>0: B_t\geq K t^{1/2} \right \} =0]=1 \quad\quad\quad(1)
$$

To prove this in Example 21.16 of Probability Theory (3rd version) A. Klenke defines $A_s:=\bigl\{\inf \left\{t>0: B_t\geq Kt^{1/2}\right\}\leq s\bigr\}$ and $A:=\bigl\{\inf \left\{t>0: B_t\geq Kt^{1/2}\right\}=0\bigr\}$ and, using the Blumenthal's 0-1 law, concludes that:

$$
P[A]=\inf_{s>0} P[A_s]\geq P[B_1\geq K]>0
$$

My questions:

1. Why is $\inf_{s>0} P[A_s]\geq P[B_1\geq K]$?
2. In a remark, the author states that this whole example shows that, for every $t\geq0$, almost surely $B$ is not Holder-$\frac{1}{2}$-continuos at $t$. How can I see it? I really struggle at this.
3. How can I interpret (1)?

Please, let me know if more context is needed. Thanks for the help.

Answer 1

1. Fix $0<r<s$ and observe that $ \{B_r\geq Kr^{1/2}\}\subset A_s$ by definition of the latter. This implies that $$\mathbb{P}(A_s) \geq \mathbb{P}(B_r \geq Kr^{1/2}) = \mathbb{P}(B_1\geq K),$$ where the last equality is due to Brownian scaling (that is, $B_r/r^{1/2}$ has the same distribution as $B_1$).
2. Recall that a function $f\colon I \to \mathbb{R}$ is Holder-continuous with exponent $1/2$ if there exists a constant $K>0$ (which I call Holder coefficient and it can always be taken to be an integer) such that $$|f(t)-f(s)|\leq K|t-s|^{1/2} \tag{2}.$$ For every integer $K\geq 1$, set $$C_K = \left\{\inf\{t>0\colon \, B_t \geq Kt^{1/2}\}=0\right\}.$$ You just showed that every $C_K$ has probability $1$. Thus, their intersection $C = \bigcap_{K\geq 1}C_K$ also has probability $1$. But what does it mean for $\omega \in \Omega$ to lie in $C$? If $\omega \in C$, then by definition, for every $K\geq 1$, we have $$\inf\{t>0\colon \, B_t(\omega) \geq Kt^{1/2}\}=0.$$ This infimum being equal to $0$ means that for any interval $[0,\epsilon]$, no matter how small $\epsilon >0$ is, you can find $t\in [0,\epsilon]$ such that $B_t(\omega) \geq K t^{1/2}$. This contradicts $(2)$ with $s=0$, implying that $B_{.}(\omega)$ is not Holder-continuous with exponent $1/2$ and coefficient $K$ on any interval containing $0$. Finally, since this holds for every $K\geq 1$, $B_{.}(\omega)$ cannot be Holder-continuous with exponent $1/2$ on any interval containing $0$.
3. The answer to your last question is above. To be more explicit, $(1)$ means that the event $C_K$ has probability $1$, that is almost surely, $B_t/t^{1/2}$ will exceed level $K$ "instantly".