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Thread: is this sequence convergent?

  1. #1
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    is this sequence convergent?



    Let Is the sequence convergent in ?
    First I tried to check if this sequence is Cauchy and I calculated integral of the sequence over . I got ). Then I'm stuck can anyone help?
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  2. #2
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    Quote Originally Posted by hermanni View Post


    Let Is the sequence convergent in ?
    First I tried to check if this sequence is Cauchy and I calculated integral of the sequence over . I got ). Then I'm stuck can anyone help?
    If the $\displaystyle L_1$-limit of the sequence $\displaystyle f_n(t) = \frac1{\sqrt n}e^{-\frac1n(t-n)^2}$ exists, then it must be equal to the pointwise limit, which is the zero function. So all you need to do is to check whether $\displaystyle \|f_n\|_1 = \displaystyle\int_0^\infty \!\! f_n(t)\,dt \to0$ as $\displaystyle n\to\infty$. In fact, if you make the substitution $\displaystyle u = \frac1{\sqrt n}(t-n)$, you see that $\displaystyle \|f_n\|_1 = \displaystyle\int_{-\sqrt n}^\infty e^{-u^2}du \to\sqrt\pi$, and so the sequence does not converge in the $\displaystyle L_1$-norm.
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    What about the same $\displaystyle x_{n} $ , is $\displaystyle { e^n x_{n} } $ is convergent in C[0,1] with sup norm? I think it converges to $\displaystyle e^{2t} $.
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  4. #4
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    Quote Originally Posted by hermanni View Post
    What about the same $\displaystyle x_{n} $ , is $\displaystyle { e^n x_{n} } $ is convergent in C[0,1] with sup norm? I think it converges to $\displaystyle e^{2t} $.
    I agree, the pointwise limit is $\displaystyle e^{2t} $. To show that this is the limit in the sup norm, you need to show that $\displaystyle e^n x_{n} \to e^{2t}$ uniformly on [0,1]. So look at the difference $\displaystyle e^n x_{n} - e^{2t}$ and see if that can be made small for all $\displaystyle t\in[0,1]$, provided that $\displaystyle n$ is large enough.
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    One more thing : can you advise good functional analysis books ? (especially on banach spaces )
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