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
Topic
Analysis
Difficulty Level
Hard
Suggested Book
Real analysis Bartle, Sherbert

Start with hints

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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