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ST the sequence $f_n$ where $f_n(x)=e^{-nx}$ is point wise but not uniformely convergent in $[0,\infty[$. Also show that the convergence is uniform in $[k,\infty[$, $k$ being a positive number.

I am able to show that point wise limit is $f(x)=0 \ \ \forall \ x$

Let $\epsilon>0$ be given then,

$|f_n(x)-f(x)|=|e^{-nx}-0|=e^{-nx}<\epsilon$

we can choose $m\in N$ such that $\large \large m>{1\over {\epsilon xlogx}} $

so as $x\rightarrow \infty \ \ \ m\rightarrow 0$ , but $m>0$ so not uniformely convergent. Is it good ?

How do i proceed further ?

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    $\begingroup$ $(f_n)$ does not converge pointwise to the zero function on $[0,\infty)$. It converges to the discontinuous function that takes the value $1$ at $x=0$ and the value $0$ elsewhere. $\endgroup$ – David Mitra Jul 19 '13 at 15:20
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The sequence $(f_n)$ is pointwise convergent on $[0,+\infty)$ to the function $f$ defined by

$$f(x)=\left\{\begin{array}\\1&\text{if}\ x=0\\ 0&\text{if}\ x>0\end{array}\right.$$ and since $f$ isn't continous then the convergence isn't uniform on $[0,+\infty)$.

For all $k>0$ we have $$\forall x\geq k,\quad|f_n(x)|\leq e^{-nk}\to_{n\to\infty}0$$ so the sequence is uniformly convergent on $[k,+\infty)$.

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  • $\begingroup$ Thanks !!, I need to be more careful with the point-wise limits in that case :) $\endgroup$ – Aman Mittal Jul 19 '13 at 15:28
  • $\begingroup$ Yes that's right:) $\endgroup$ – user63181 Jul 19 '13 at 15:29
  • $\begingroup$ 1 side question, If $f_n(x)=tan^{-1}nx, x\ge 0$ what will be its point wise limit in $[0,b]$ ?? $\endgroup$ – Aman Mittal Jul 19 '13 at 16:22
  • $\begingroup$ The limit $f$ is defined by: $f(0)=0$ and $f(x)=\pi/2$ if $x>0$. $\endgroup$ – user63181 Jul 19 '13 at 16:25
  • $\begingroup$ Can you please elaborate on how $f(x)=\pi /2$ for $x>0$ ? I am missing out on some concept here i think $\endgroup$ – Aman Mittal Jul 19 '13 at 16:26

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