We've got a sequence of functions
$f_n(x) = \frac {x}{nx^2 +4}$
Some questions:
- Determine the pointwise limit of $(f_n)$. Call this function $f$
- Compute the maximum and minimum of each function $f_n$. Prove now that the convergence $f_n \rightarrow f$ is uniform
- Determine the sequence $(f'_n)$. For which $x$ does $f'_n(x)$ converge to $f'(x)$?
- Considering the following theorem, are the conditions satisfied? :
Let $f_n \rightarrow f$ pointwise on the closed interval [a,b], and assume that each $f_n$ is differentiable. If $(f'_n)$ converges uniformly on [a,b] to a function g, then the function $f$ is differentiable and $f'=g$
My try
- $\lim\limits_{n \to \infty}f_n(x) = \frac{x}{\infty}=0 $
- Using the Quotient Rule, we have $ f'_n(x)= \frac {nx^2+4-2nx^2}{(nx^2+4)^2} = 0$ Looking for extrema we set $nx^2+4-2nx^2 =0 \implies -nx^2 +4 = 0 \implies x = \pm \sqrt{\frac {n}{4}}$. We have a minimum on $x=-\sqrt{\frac{n}{4}}$ and a maximum on $x=\sqrt{\frac{n}{4}}$ for all $n$. The values of the extreme are: for the minimum $f_{n,min}=f_n(-\sqrt{\frac{n}{4}}) = \frac{-\sqrt{{\frac{n}{4}}}}{n\frac{n}{4} +4} $ $f_{n, max}= f_n(\sqrt{\frac{n}{4}})= \frac{\sqrt{{\frac{n}{4}}}}{n\frac{n}{4} +4}$ So, now we know that $f_n$ is bounded between the minimum and maximum value. How can I proof that the convergence is uniform?
- I don't understand
- I have to check wheter $(f'_n)$ converges uniformly to a function g, and I have to check wheter $f$ is differentiable?