Answer to Question #170281 in Real Analysis for Prathibha Rose

Question

Let f_n(x) =1/(1+n²x²) if 0 less than or equal to x less than or equal to 1,n=1,2,........ Prove that {f_n} converges point wise ,but not uniformly on [0,1] Is term by term integration permissible?

Answer 1

Here 

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

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

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

Let us take the interval [0,1] containg 0.

Thus

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

     $= \underset{x\in [a,b]}{Sup}|\dfrac{1}{1+n^2x^2}|$

     $=\dfrac{1}{2}$ Which does not tends to zero as $n\rightarrow \infty.$

 Hence The sequence is not uniformly convergent in interval [0,1].