# Uniform convergence and derivatives

For $n=1,2,3,\dots,$ and $x$ real, put $f_n(x)=\frac{x}{1+nx^2}$. Can you help me to show that $(f_n)$ converges uniformly to a function $f$, and that the equation $\lim(f'_n(x))= f'(x)$ is not correct if $x=0$, but true if $x$ is not equal $0$?

• I tried to show that the function fn is uniformly convergent to 0 – birzh May 1 '13 at 19:16
• According to my calculations, your final statement is backwards: the limit is invalid for $x = 0$, valid for $x \ne 0$. – TonyK May 1 '13 at 19:17
• yes you are right' – birzh May 1 '13 at 19:21
• – Mhenni Benghorbal May 1 '13 at 19:53

Hint

It's pretty clear that $(f_n)$ is pointwise convergent to the zero function $f$ on $\mathbb{R}$ and we have $$f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}=0\iff x=\frac{1}{\sqrt{n}}=x_n$$ and since $$||f_n-f||_\infty=|f_n(x_n)|=\frac{1}{2x_n}\to0$$ so $(f_n)$ is uniformly convergent to $f$ on $\mathbb{R}$

Finally we can see easily that for $x=0$, $f'_n(0)=1\to 1\neq f'(0)=0$ and for $x\neq 0$ $f'_n(x)\to 0=f'(x)$

Let's prove that $f_n$ converges uniformly to 0 in $\mathbb R$.

From the definition:

$$\sup\limits_{x\in\mathbb R}|f_n(x)-0|=\sup\limits_{x\in\mathbb R}\dfrac{|x|}{|1+nx^2|}=$$ $$\sup\limits_{x\in\mathbb R}\dfrac{|x|}{1+\sqrt{n}|x|^2}\leq\sup\limits_{x\in\mathbb R}\dfrac{|x|}{2\sqrt{n}|x|}=\dfrac{1}{2\sqrt{n}}\overset{n\rightarrow\infty}{\longrightarrow }0$$ Therefore, $\sup\limits_{x\in\mathbb R}|f_n(x)-0|\rightarrow 0,$ i.e. $f_n$ converges uniformly to $0$.

Now, take $f_n'(x)=\dfrac{1-nx^2}{(1+nx^2)^2}$

• If $x=0$ then $f_n'(0)=1\nrightarrow f'(0)=0$
• If $x\neq 0$ then you can easily check that $f_n'(x)\rightarrow f'(x)=0$

Since you have to calculate derivatives anyway, you might as well apply the standard method from calculus (argue about the sign of $f'_n$ on three intervals) to show that $\max (|f_n(x)| : x\in \mathbb{R})=\dfrac{1}{2\sqrt{n}}$ and argue from there about uniform convergence. Obviously, you want to show that $f'_n(x)\to 0$ if $x\neq 0$ and $f'_n(0)$ does not. The nice thing about $f_n'(0)$ is that it's a constant sequence.