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Let $f_n(x)=e^{-nx}$ on the interval [0,1]. Explain why the sequence of functions {$f_n$} converges pointwise on [0,1]. What is the limiting function and is it continuous? Is the convergence uniform?

I think I have an intuitive idea of how to do this, but formalizing it and the portion on uniform convergence trip me up.

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  • $\begingroup$ Can you find the pointwise limit? $\endgroup$ – Antonio Vargas Oct 26 '15 at 4:12
  • $\begingroup$ Would the pointwise limit just be e--when x=0? $\endgroup$ – byong Oct 26 '15 at 4:46
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To prove that $f(x)=e^{-nx}$ converges pointwise, pick a fixed x first. If $x = 0$, $f(x)=1$ as $n \rightarrow \infty$. If $x>0$, $f(x)=0$ as $n \rightarrow \infty$. So $f(x)$ does converges pointwise. Now the limiting function is not continuous at $0$ as you can see.

The idea of uniformly convergent is that your convergence does not depend on $x$. As far as the intuition goes, think about they gave you any large number $N$, as you pick $x$ close enough to zero, $f_n(x)$ will approach $1$ eventually as $x$ gets close enough to $0$, which makes it impossible to make $|f_n(x)-f(x)|<\epsilon$

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  • $\begingroup$ I agree with the first paragraph, but not the second is obtuse. To me uniform convergence would imply that f(1) - f(2) = f(2) - f(3) = f(3) - f(4) and so on. That obviously isn't so since you would be summing a finite quantity an infinite number of times. Such a sum would be unbounded. $\endgroup$ – MaxW Oct 26 '15 at 7:24

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