 # Understand the problem

Let $\{f_n\}$ be a sequence of functions from $\Bbb R$ to $\Bbb R$ defined by $f_n(x)=\frac 1n exp(-n^2x^2)$ Then which one of the following statements is true?
1. Both the sequences $\{f_n\}$ and $\{f_n'\}$ converge uniformly on $\Bbb R$.
2. Neither $\{f_n\}$ nor $\{f_n'\}$ converges uniformly on $\Bbb R$.
3. $\{f_n\}$ converges pointwise but not uniformly on any interval containing the origin
4. $\{f_n'\}$ converges pointwise but not uniformly on any interval containing the origin.
##### Source of the problem
TIFR 2019 GS Part A, Problem 12
Analysis
Hard
##### Suggested Book
Real analysis Bartle, Sherbert

Do you really need a hint? Try it first!
Check the function is uniformly continuous or not using sup-norm definition i.e $M_n=Sup_{x\in I}|f_n(x)-f(x)|$ then the function is uniformly continuous if $M_n \to 0$ as $n \to \infty$ $M_n=Sup_{x\in \Bbb R}\frac 1n exp(-n^2x^2)=\frac 1n$ so $M_n \to 0$ as $n \to \infty$ then function is uniformly continuous

Now, $f_n'=-2nx exp(-n^2x^2) \to 0$ as $n \to \infty$ in $\Bbb R$. Consider $0\subseteq I \subseteq \Bbb R$. Calculate $M_n=Sup_{x\in I}|f_n'(x)|$.  What can you say about option 4? $M_n=Sup_{x\in I}|f_n'(x)|>|f_n'(\frac 1n)|=2e^{-1}$.

Can you comment about $f_n'(x)$?

It is clear that $latex f_n'(x)$ is not uniformly convergent on any such $latex I$. So option 4) is correct

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