# uniform convergence on [0, 1]

determine whether $f_n (x) = \frac {x} {1+nx^2}$ defines the sequence of functions $\{f_n\} _{n\in N}$ converging uniformly on [0, 1]?

My take on this question:

First, I determined the pointwise on this function which is ${f_n} \to 0$ as $n \to \infty$ and then I found $M_n = sup{|f_n (x) - f(x)|} = \frac {1/ \sqrt(n)}{1+n/\sqrt n}$. Therefore the sequence of functions converge to [0,1] uniformly as $M_n \to 0$ as $n \to \infty$

I am not sure if my answer and/or approach is correct. If someone can please verify for me, please?

Thank You in advance!!!

## 1 Answer

The numerator of $f'_n (x)$ is

$$1+nx^2-2nx^2=1-nx^2$$

the maximum $M_n$ is attained at $x=1/\sqrt {n} .$

but $$M_n=\frac {1}{2\sqrt {n}}\to 0$$ thus the convergence is uniform at $[0,1]$.

• just a side note/question: max $M_n$ is attained at the highest x value alwaya, right? – ISuckAtMathPleaseHELPME Dec 5 '17 at 22:27
• @ISuckAtMathPleaseHELPME The extremum is attained where the derivative is zero. – hamam_Abdallah Dec 5 '17 at 22:35