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Thread: uniform convergence

  1. #1
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    uniform convergence

    Hi,

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

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

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

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

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

    Can anybody help me?
    Thanks!
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  2. #2
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    Quote Originally Posted by Juju View Post
    Hi,

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

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

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

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

    Does $\displaystyle (f_h)_{h \in [0,1]}$ also converge uniformly to $\displaystyle f$? 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.
    ..
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  3. #3
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    Thanks!

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

    How can I show the uniform convergence under this condition?
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  4. #4
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    Quote Originally Posted by Juju View Post
    Thanks!

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

    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 $\displaystyle (1+th)^{1/h}$ increase monotonically to $\displaystyle e^t$ as $\displaystyle h\searrow0$, 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 $\displaystyle \varepsilon>0$ there exists $\displaystyle h_\varepsilon$ with $\displaystyle 0<h_\varepsilon<1$ such that $\displaystyle |(1+th)^{1/h} - e^t|<\varepsilon$ for all $\displaystyle t\in[0,T]$ whenever $\displaystyle 0<h<h_\varepsilon.$

    Next, the function $\displaystyle (a,h)\mapsto g(a,h) \mathrel{\overset{\text{d{e}f}}{=}}a(\mu-r)+\frac{1}{2}a^2 \gamma(1-\gamma)\sigma^2+Ch^\frac{1}{2}$ is continuous and hence bounded when $\displaystyle a\in[0,1]$ and $\displaystyle h\in[0,1]$, say $\displaystyle |g(a,h)|\leqslant T.$ Hence $\displaystyle |(1+g(a,h)h)^{1/h} - e^{g(a,h)}|<\varepsilon$ whenever $\displaystyle 0<h<h_\varepsilon.$ Since g(a,h) is close to g(a,0) for h sufficiently small, that should be enough to show that $\displaystyle f_h(a)$ converges to $\displaystyle e^{g(a,0)}$ uniformly on [0,1].
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  5. #5
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    Sorry, my thread wasn't true...
    Last edited by Juju; Mar 21st 2011 at 06:12 AM.
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