Uniform convergence and Integrals

Question

For $x \in \mathbb{R}$ and $n \in \mathbb{N}$, let $f_n(x) = \frac{x}{1+nx^2}$ and $f(x)=0$.

(i) Prove $f_n \to f$ uniformly.

(ii) Calculate $\displaystyle{\lim_{n\to\infty}\int_0^1} f_n(x)dx$ in $2$ ways.

(iii) For what values of $x\in\mathbb{R}$ is it true that $\displaystyle{\lim_{n\to\infty}} f'_n(x) = f'(x)?$

For (i) I set $M_n = sup|f_n(x)-0| < \frac{1}{x}$. Since $\displaystyle{\lim_{x\to\infty}} \frac{1}{x} \to 0$ the convergence is uniform. (I am not positive this is correct)

For (ii) we proved we can take the limit inside the integral, so we get $\displaystyle{\int_0^1} f(x) dx= 0$. I directly did the integral then took the limit and once again got $0$. Is this legal to do? I cannot think of any other way to do this.

For (iii) The left hand side is $0$ so then when is $\frac{1-x^2}{x^4+2x^2+1} = 0$. This is only true at $x=1,-1$.

Answer 1

(i) is incorrect: you did not understand how to prove uniform convergence.

$$ M_n = \sup_x \frac{|x|}{1 + nx^2} $$ You want to prove that this goes to 0 when $n\to\infty$.

We have : $$ \frac{|x|}{1 + nx^2} \le \frac{|x|}{ nx^2} = \frac 1 {n|x|} $$but it is not small when $x$ is close to 0.

Let $r>0$.

• For $x$ small, the numerator is small: $$ \frac{|x|}{1 + nx^2} \le |x| \le r $$ when $|x| \le r$.
• Otherwise, $$ \frac{|x|}{1 + nx^2} \le \frac 1 {n|x|} \le \frac 1 {nr} \le \frac 1 r $$ as soon as $n> \frac 1 {r^2}$.

$$ n> \frac 1 {r^2} \Rightarrow M_n \le r $$ that is $$M_n\to_{n\to\infty} 0 $$

(ii) is correct.

(iii) is not, there is a mistake in your computation (where is $n$?). See the edit by OP in the comments for the solution.