I would like to prove that the following random variables converge in distribution: $X_h(t)=\frac{1}{h}(B(t+h)-B(t))$ where $(B(t):t\geq0)$ is standard Brownian Motion.
In my attempts at this I used the definition that: $\lim_{h \to +\infty}\mathop{\mathbb{E}}(f(X_h))=\mathop{\mathbb{E}}(f(X))$ for all continuous bounded functions $f$.
My thoughts so far are that the process $X$ will be equal to 0 (though I haven't proved pointwise convergence either) so $\mathop{\mathbb{E}}(f(X))$ will just be a constant value.
I can also see that $X_h(t)\stackrel{}{\sim}N(0,\frac{1}{h})$. Would it then suffice to show that $F_h(t)\rightarrow F(t)$ where $F_h(t)$ is the cumulative distribution function for $X_h$ and $F(t)=0$? If so, how is this done?
EDIT: Corrected the definition of the process I am looking at, previously was $X_h(t)=\frac{1}{h}(B(t+h)-B(h))$
$EX^{2}_h(t)=\frac 1 {h^{2}} var (B_{t+h}-B_h)=\frac t {h^{2}} \to 0$ as $ h \to \infty$. So $X_h(t) \to 0$ in mean square. This is much stronger than convergence in distribution. (It implies convergence in probability).