Hey everyone I'm currently working on some properties of the Kolmogorov distribution, which we have defined as
$F\left(x\right):=\mathbb{P}\left(\sup_{t \in\left[0,1\right]}|B^{0,1}_t|\le x\right)=\begin{cases}1+2∑_{j=1}^∞ \left(-1\right)^j\exp \left(-2j^2x^2\right),x>0\\0,x\le 0\end{cases}$
Now the task is to show that:
(a) $F(x)\le \exp(-2x^2)$ for $x \in (0,1/2)$
(b) $\forall\epsilon\in(0,1/2)$ $\exists N \in \mathbb{N}:$ $\forall n\ge N$ $\int_{0}^{1/2}(1-(F(\sqrt{x\log(n})))^n)dx \ge1/2-\epsilon$
To $(a)$ clearly we have an alternating real null sequence so by Leibnitz criteria for alterning series we now it converges. If we take partial sums we have the following relation:
$∑_{j=1}^{2k+1}\left(-1\right)^j\exp \left(-2j^2x^2\right)\le ∑_{j=1}^{∞}\left(-1\right)^j\exp \left(-2j^2x^2\right)\le∑_{j=1}^{2k+2}\left(-1\right)^j\exp \left(-2j^2x^2\right)$
and for $k=0$ we find
$F(x)\ge 1-2\exp(-2x^2)$ for $x>0$ and $F(x)\le1-2\exp(-2x^2)+2\exp(-8x^2)$
The second expression on the right hand side is clearly bounded by $\exp(-2x^2)$ if $x\in (0,1/2)$ but we only have strict inequality then because equality only arises in the limit $x\to0$. So my question if there is any other way to show that $F(x)\le \exp(-2x^2)$ for $x \in (0,1/2)$ ?
Thank you in advance for any hints.