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I want to show if the following sequence of functions converges pointwise and converges uniformly on R? $ f_n(x) = e^{-nx^2}. $

Here's what I did: -Fix $\displaystyle x\in R, \lim_{n\rightarrow\infty } f_n(x) = \lim_{n\rightarrow\infty }e^{-nx^2}=0 $, so $f_n $ converges point wisely to $f $ s.t. $f(x) = 0.$ Now for uniform convergence: $\displaystyle \|f_n - f\|_{\sup} = \sup_{x\in R}\left|e^{-nx^2}-0\right| $. Now how do I show that $\displaystyle \sup_{x\in R}\left|e^{-nx^2}\right| $ goes to $0$ as $n\rightarrow \infty $ and is thus uniformly convergent?

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    $\begingroup$ The sequence is not uniformly convergent; the limit is discontinuous. Your first line is also incorrect - you've missed a case. $\endgroup$ – T. Bongers Feb 28 '18 at 17:53
  • $\begingroup$ is it the case when x=0? Then f(x) = 1 if x=0 and 0 if x is not 0. so it is pointwise convergent right? $\endgroup$ – Jack Feb 28 '18 at 17:54
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    $\begingroup$ Yes, it is. ${}$ $\endgroup$ – T. Bongers Feb 28 '18 at 17:57
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As you've noted in your question and the comments, we have two cases:

  • If $x \ne 0$, then the sequence is of the form $(e^{-x^2})^n$ which is a decaying exponential.

  • If $x = 0$, the sequence of values is constant $1$.

Hence, we have convergence everywhere. The limit is discontinuous (since it is $1$ at $0$ and $0$ otherwise), so the convergence is non-uniform.

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  • $\begingroup$ When showing that it's not uniformly convergent: I have to prove $\|f_n - f\| \nrightarrow 0 $. How do you go about this? $\endgroup$ – Jack Feb 28 '18 at 18:20
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    $\begingroup$ By proving that $\|f_n - f\| = 1$ for all $n$. $\endgroup$ – T. Bongers Feb 28 '18 at 19:11
  • $\begingroup$ Is this at x=0? $\endgroup$ – Jack Feb 28 '18 at 19:12

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