Consider a branching brownian motion i.e. we start with one particle in $0$ which evolves as a brownian motion and after a exponential time of mean $1$ splits into two particles that evolve independently from each other as their father. Let $x(t)$ be te rightmost particle at time $t$ and $T_n$ be the $n$th splitting time. I want to prove that $$\frac{\mathbb{E}(x(T_n))}{\mathbb{E}(T_n)} \to \sqrt{2}$$ Following the arguments in this paper (lemma 5.3. page 27) we know that
$$\frac{x(t)}{t} \stackrel{t\to\infty}{\longrightarrow} \sqrt{2}\quad \text{ a.e.}$$ and since $T_n \to \infty$ a.e. we have that $$\frac{x(T_n)}{T_n} \stackrel{n\to\infty}{\longrightarrow} \sqrt{2} \quad \text{ a.e.}$$ It is also easy to show that $$\frac{T_n}{\mathbb{E}(T_n)} \to 1 \quad \text{a.e.}$$ so $$\frac{x(T_n)}{\mathbb{E}(T_n)} \to \sqrt{2} \quad \text{a.e.}$$ Now an argument of dominated convergence is needed so we need to prove that $$\mathbb{E}\left(\sup_{n}\frac{|x(T_n)|}{\mathbb{E}(T_n)}\right)<\infty$$ In the paper they prove two things $$\mathbb{E}\left(\sup_{t\geq1} \frac{|x(t)|}{t}\right)^2 < \infty$$ and $$\mathbb{E}\left(\sup_{n} \frac{T_n}{\mathbb{E}(T_n)}\right)^2 < \infty$$ The second one is donde in the paper but the first one don't translate well in this context.
If anyone has an idea how to prove it or how to get the same result with another technique it will be great.
Any help will be appreciated