Given the function sequence below:
$$f_n(x)=e^{-x^2/n}, \quad x \in \Bbb R$$
Is the functional sequence $f_n(x)$ convergent uniformly?
I know that $f_n(x)$ is convergent to $f(x)=1$ point wise, but I cannot show uniformly convergent of this.
Uniform convergence of function sequence
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real-analysis
sequences-and-series
convergence
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1You should add some context and describe where you got stuck. Questions in this form are easy targets for flags, downvote, closure etc. – 2017-02-25
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0You mean in $C(\Bbb R)$? The pointwise limit is zero everywhere except $x = 0$, so it's not continuous, so it's not a uniform convergence under the sup-norm. If we get rid of this point, then the answer is positive. – 2017-02-25
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0@GNUSupporter Why did you exclude $x=0$? – 2017-02-25
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0@polfosol I've confused $x^2/n$ with $nx^2$ – 2017-02-25
2 Answers
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I hope you know that if a sequence of continuous function converges uniformly, the the following holds: \begin{align} \lim\limits_{x\to a} \lim\limits_{n\to \infty}f_n(x)=\lim\limits_{{n\to \infty}} \lim\limits_{x\to a}f_n(x)\end{align} If that does not hold, then the function does not converge uniformly. Now look at this: \begin{align} \lim\limits_{x\to \infty} \lim\limits_{n\to \infty}f_n(x) = \lim\limits_{x\to \infty} 1 =1\end{align} But on the other hand we have: \begin{align} \lim\limits_{{n\to \infty}} \lim\limits_{x\to \infty}f_n(x) =\lim\limits_{{n\to \infty}} \lim\limits_{x\to \infty} e^{\frac{-x^2}{n}} =\lim\limits_{{n\to \infty}} 0 = 0 \neq 1 = \lim\limits_{x\to \infty} \lim\limits_{n\to \infty}f_n(x) \end{align} Thus, the sequence doesn't converge uniformly.
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This sequence of functions is not uniformly convergent.
To see this, suppose the sequence $f_n(x)$ is uniformly convergent. Then it would have to converge to $1$, since uniform convergence implies pointwise convergence, and you have identified that the sequence converges pointwise to $1$.
And if $f_n(x)$ uniformly converges to $1$, then by definition, for every $\epsilon > 0$, there exists an $N \in \mathbb N $ such that $$ n > N \implies \sup_{x \in \mathbb R} | e^{-x^2/n} - 1 | < \epsilon.$$
Are you able to determine $\sup_{x \in \mathbb R} | e^{-x^2/n} - 1 | $, for a given $n$?
Can you find an $\epsilon$ such that the above inequality fails? Good luck!