Let $B_t$ be a Brownian motion, $B_T^* = \sup_{0\leq t \leq T} B_t$ and $\lambda > 0$. Applying Doob's maximal inequality gives: \begin{align} P(B_T^* \geq \lambda)\leq \frac{\mathbb{E}[B_T^p]}{\lambda^p}=\frac{T}{\lambda^2},\qquad \text{for $p=2$}. \end{align} By applying the convex function $x \mapsto (x+c)^2$ for $B_t$ for a suitable constant $c$, I want to find that \begin{align} P(B_T^* \geq \lambda)\leq \frac{T}{\lambda^2 +T}. \end{align} However, for me it is unclear how to use correctly this convex function. Since, $B_t \mapsto (B_t+c)^2$ Jensens' inequality yields that \begin{align} \mathbb{E}[(B_t + c)^2] &\geq (\mathbb{E}[B_t] + c)^2 \\ \mathbb{E}[B_t^2] + 2c\ \mathbb{E}[B_t] + c^2 &\geq c^2 \\ t &\geq 0, \end{align} which is trivial for every constant $c$.


First of all, note that

$$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \mathbb{P}\left( \sup_{t \in [0,T]} (B_t+c)^2 \geq (\lambda+c)^2 \right).$$

Since $M_t := (B_t+c)^2$ is a non-negative submartingale, it follows from Doob's maximal inequality that

$$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \frac{1}{(\lambda+c)^2} \mathbb{E}((B_T+c)^2).$$

As $\mathbb{E}(B_T^2)=T$ and $\mathbb{E}(B_T)=0$, this shows

$$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \frac{T+c^2}{(\lambda+c)^2}. \tag{1}$$

It remains to choose the constant $c$ in such a way that the right-hand side gets as small as possible. To this end, define

$$g(c) := \frac{T+c^2}{(\lambda+c)^2}$$


$$g'(c) = \frac{2c}{(\lambda+c)^2} -2 \frac{(T+c^2)}{(\lambda+c)^3} = \frac{1}{(\lambda+c)^3} (2c(\lambda+c)-2(T+c^2))$$

equals $0$ if

$$c = \frac{T}{\lambda}.$$

Plugging this particular $c$ into $(1)$, we get

$$\mathbb{P}(B_T^{\ast} \geq \lambda) \leq \frac{T+\frac{T^2}{\lambda^2}}{(\lambda+\frac{T}{\lambda})^2} = \frac{T}{\lambda^2} \frac{1+ \frac{T}{\lambda^2}}{\left(1+ \frac{T}{\lambda^2} \right)^2} = \frac{\lambda}{T^2} \frac{1}{1+\frac{T}{\lambda^2}} = \frac{T}{\lambda^2+T}.$$


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