Question

Let F n(x) =1/n *e^(-n^2*x^2) if x element of R ,n=1,2 .....

Prove that fn tend to 0 uniformly on R ,that derivative of fn tend to 0 point wise on R ,but that the convergence of {fn'} is not uniform on any interval containing the origin

Answer 1

Here 

$\underset{{n\rightarrow \infty}}{lim }f_n(x)=0, \forall x$

And $f_n(x)=\dfrac{e^{-n^2x^2}}{n}$

$f'_n(x)=\dfrac{-2xn^2 e^{-n^2x^2}}{n}$

so $f'_n(x)=-2xne^{-n^2x^2}$

Now $-2xne^{-n^2x^2}$ attains maximum value $\dfrac{2}{e} \text{ at } x=\dfrac{-1}{n}$

 tending to 0 as $n\rightarrow \infty.$

Let us take the interval [a,b] containg 0.

Thus

  $M_n=\underset {x\in [a,b]}{Sup}|f_n(x)-f(x)|$

  $= \underset{x\in [a,b]}{Sup}|-2xne^{-n^2x^2}|$

     $=\dfrac{2}{e}$ Which does not tends to zero as $n\rightarrow \infty.$

 Hence the sequence ${f_n}$ is not uniformly convergent in any interval [a,b] containing the origin.