# Math Help - uniform convergence

1. ## uniform convergence

Hi,

is the following sequence of functions $(f_h)_{h \in [0,1]}$ uniformly convergent for $h \to 0$?

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

whereas
$\gamma \in (0,1)$ is a constant
$C, \mu \in \mathbb{R}$ are constants
$\sigma,r >0$ are constants

I think the pointwise limit is given by
$\lim\limits_{h \to 0}f_h(a)=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}.$
Is this correct?

Does $(f_h)_{h \in [0,1]}$ also converge uniformly to $f$?

Can anybody help me?
Thanks!

2. Originally Posted by Juju
Hi,

is the following sequence of functions $(f_h)_{h \in [0,1]}$ uniformly convergent for $h \to 0$?

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

whereas
$\gamma \in (0,1)$ is a constant
$C, \mu \in \mathbb{R}$ are constants
$\sigma,r >0$ are constants

I think the pointwise limit is given by
$\lim\limits_{h \to 0}f_h(a)=e^{a(\mu-r)+ \frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2}.$
Is this correct? Yes.

Does $(f_h)_{h \in [0,1]}$ also converge uniformly to $f$? Uniform convergence is only defined if the domain of the functions is specified. If you want the variable a to run through the whole real line, then the convergence is not uniform. But I think that the functions will converge uniformly on any bounded interval.
..

3. Thanks!

Sorry, I' ve forgotten to mention that $a \in [0,1]$.

How can I show the uniform convergence under this condition?

4. Originally Posted by Juju
Thanks!

Sorry, I' ve forgotten to mention that $a \in [0,1]$.

How can I show the uniform convergence under this condition?
I was hoping you wouldn't ask that, because it doesn't seem to be obvious. Here's one approach that appears to work.

First, the functions $(1+th)^{1/h}$ increase monotonically to $e^t$ as $h\searrow0$, and it follows from Dini's theorem that the convergence is uniform on any finite interval [0,T] (where T is to be chosen in the next paragraph). Thus given $\varepsilon>0$ there exists $h_\varepsilon$ with $0 such that $|(1+th)^{1/h} - e^t|<\varepsilon$ for all $t\in[0,T]$ whenever $0

Next, the function $(a,h)\mapsto g(a,h) \mathrel{\overset{\text{d{e}f}}{=}}a(\mu-r)+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2+Ch^\frac{1}{2}$ is continuous and hence bounded when $a\in[0,1]$ and $h\in[0,1]$, say $|g(a,h)|\leqslant T.$ Hence $|(1+g(a,h)h)^{1/h} - e^{g(a,h)}|<\varepsilon$ whenever $0 Since g(a,h) is close to g(a,0) for h sufficiently small, that should be enough to show that $f_h(a)$ converges to $e^{g(a,0)}$ uniformly on [0,1].

5. Sorry, my thread wasn't true...