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Math Help - uniform convergence

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Juju View Post
    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.
    ..
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  3. #3
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    Thanks!

    Sorry, I' ve forgotten to mention that 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 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<h_\varepsilon<1 such that |(1+th)^{1/h} - e^t|<\varepsilon for all t\in[0,T] whenever 0<h<h_\varepsilon.

    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<h<h_\varepsilon. 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].
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  5. #5
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    Sorry, my thread wasn't true...
    Last edited by Juju; March 21st 2011 at 07:12 AM.
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