Consider an SBM $(B_t)_{t\geq 0}$. Now we can obtain a polygonal path on $[0,n]$ by joining the integral points $B_0, B_1, \ldots, B_n$ with segments and call this path $B^{n}_t$. Now I want to bound $\sup\limits_{0 \leq t \leq n}|B_t - B^{n}_t|$ by a poly-logarithmic bound with high concentration, i.e. something with the following flavour:
$$\mathbb{P}\big\{\sup\limits_{0 \leq t \leq n}|B_t - B^{n}_t| \geq A(\log n)^p\big\} \leq \frac{\alpha}{n}$$ , where $A, p$ and $\alpha$ are positive constants. It seems it should be true since $\sup\limits_{0 \leq t \leq 1}(B_{n+t} - B_n)$ (similarly infimum) is distributed as $|N(0,1)|$ and are independent for different $n$'s. And asymptotic growth rate of maximum of i.i.d. standard normals is like $\sqrt{\log n}$. But I want a concrete result. Any suggestions?
Clearly,
$$\begin{align*} \sup_{0 \leq t \leq n} |B_t-B_t^n| &= \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]} |(1-t) \cdot (B_{t+k}-B_k)+t \cdot (B_{t+k}-B_{k+1})| \\ &\leq \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]}|B_{t+k}-B_k| + \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]} |B_{k+1}-B_{(k+1)-t}|. \end{align*}$$
Note that $(B_t^k)_{t \in [0,1]} :=(B_{t+k}-B_k)_{t \in [0,1]}$ as well as $(W_t^k)_{t \in [0,1]} := (B_{k+1}-B_{(k+1)-t})_{t \in [0,1]}$ are Brownian motions. Consequently,
$$\mathbb{P} \left( \sup_{t \leq n} |B_t-B_t^n| \geq r \right) \leq 2 \mathbb{P} \left( \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]} |B_{t+k}-B_k| \geq \frac{r}{2} \right).$$
Moreover, using that $(-B_t)_{t \geq 0}$ is a Brownian motion as well, we conclude that it suffices to estimate probabilities of the form
$$I:=\mathbb{P} \left( \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]} (B_{t+k}-B_k) \geq r \right) \tag{1} $$
By the reflection principle, we have $\sup_{t \in [0,1]} (B_{t+k}-B_k) \sim |B_1|$. Consequently,
$$I \leq n \cdot \mathbb{P}(|B_1| \geq r) \leq \frac{n}{\sqrt{2\pi}} \frac{1}{r} \exp \left(-\frac{r^2}{2} \right) \tag{2}$$
for any $r>0$. In particular, for $r =\sqrt{2A \log n}$,
$$I \leq \frac{n}{A \sqrt{2\pi}} \frac{n^{-A}}{\log n}$$
If we choose $A = 2$, this proves in particular
$$\mathbb{P} \left( \max_{k=0,\ldots,n-1} \sup_{t \in [0,1]} (B_{t+k}-B_k) \geq \sqrt{4 \log n} \right) \leq \frac{1}{2 \sqrt{2\pi}} \frac{1}{\log n \cdot n} \tag{3}$$
In fact we can replace $\sqrt{2A \log n}$ by $(A \log n)^p$ for some $p>0$, but this makes the estimates more messy; in particular we have to modify $A$ in order to obtain an estimate of the form $(3)$.