Hallo,

Someone already helped me to show that the following convergence is uniformly for all $\displaystyle a \in [0,1]$:

$\displaystyle \lim\limits_{h \to 0}\left(1+a(\mu-r)h+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2h+O(h^\frac{3}{2})\right)^\frac{1}{ h}=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}.$

whereas

$\displaystyle O(h^\frac{3}{2})$ is the remainder term of a taylor expansion where all terms with factor $\displaystyle h^\frac{3}{2}$ and higher exponents are collected. This term can be bounded by a constant that can be chosen independent of $\displaystyle a$

$\displaystyle \mu \in \mathbb{R}$ is a constants

$\displaystyle \sigma,r >0 $ are constants

$\displaystyle 0<\gamma <1 $ is a constant

$\displaystyle f_h(a):=\left(1+a(\mu-r)h+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2h+Ch^\frac{3}{2})\right)^\frac{1}{h }$

$\displaystyle f(a):=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}$

does the maximum point of the sequence of functions $\displaystyle (f_h(a))_{h \in (0,1)}$ (I call it $\displaystyle x_h$) with $\displaystyle a \in [0,1]$ converge to the maximum point of $\displaystyle f$ for (I call it $\displaystyle x$) $\displaystyle h\to 0$?

Does anyone know if the following holds:

$\displaystyle \lim\limits_{h \to 0}x_h = x$?

I only know a theorem which says that this is true for a sequence of concave functions. But my functions are neither concave nor convex for all $\displaystyle a \in [0,1]$.

Can anybody help me?

Thanks!