Let $B$ be a standart Brownian motion, define $K:=\{t\in [0,1]: B_t=0\}$. I want to show that for all $0\leq m\leq n-1$,$$\Bbb{P}\left(K\cap \left[\frac{m}{n}, \frac{m+1}{n}\right]\neq\emptyset\right)\leq \Bbb{P}\left(\sup_{t\in\left[0,\frac{1}{n}\right]}\big|B_{\frac{m}{n}+t}-B_{\frac{m}{n}}\big|\geq\big|B_{\frac{m}{n}}\big|\right)~~~~~~~~~~~~~~~~~~(1)$$
I know that $$K\cap \left[\frac{m}{n}, \frac{m+1}{n}\right]\neq\emptyset~~~\Leftrightarrow~~~ \exists t_0\in \left[\frac{m}{n}, \frac{m+1}{n}\right]: B_{t_0}=0$$ Using this we see that $\big|B_{t_0}-B_{\frac{m}{n}}\big|=\big|B_{\frac{m}{n}}\big|$ since $t_0 \in \left[\frac{m}{n}, \frac{m+1}{n}\right]$ we know that there exists $\xi \in \left[0,\frac{1}{n}\right]$ s.t. $t_0=\frac{m}{n}+\xi$. Hence $\sup_{t\in\left[0,\frac{1}{n}\right]}\big|B_{\frac{m}{n}+t}-B_{\frac{m}{n}}\big|\geq\big|B_{\frac{m}{n}}\big|$. But is this enough to show my inequality $(1)$?
Thanks for your help.
Yes. You just showed that $$ \left\{K\cap \left[\frac{m}{n}, \frac{m+1}{n}\right]\neq\emptyset\right\}\subset\left\{\sup_{t\in\left[0,\frac{1}{n}\right]}\big|B_{\frac{m}{n}+t}-B_{\frac{m}{n}}\big|\geq\big|B_{\frac{m}{n}}\big|\right\}, $$
so the probability of the left-hand side must be smaller than or equal to that of the right-hand side.