Is the function uniformly continuous? .

Question

$f_n(x)= \dfrac{nx}{1+n^4x^4}$ , $x∈\mathbb{R}$

I fixed $\varepsilon >0$, looking for $\delta >0$, so for every $x, y ∈ \mho$ with $|x-y|<\delta$ it is valid $|f(x)-f(y)|<\epsilon$.

So: $\left|\dfrac{nx}{1+n^4x^4}-\dfrac{ny}{1+n^4y^4}\right|=\left|\dfrac{n(x-y)+n^5xy(y^3-x^3)}{(1+n^4x^4)(1+n^4y^4)}\right|$

I don't know how can I go on to find a $\delta$, is something wrong above?

Answer 1

You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $d\left|\dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}\right|$. So now if you can prove there is some constant number $M$ such that $\left|\dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}\right|<M$ for all $x$, $y$, you can take $\delta=\epsilon/M$. Then if $|d|=|y-x|<\epsilon/M$, you have $|f(y)-f(x)|<dM=\frac{\epsilon M}M=\epsilon$.

Answer 2

The derivative of $f_1$ is bounded: you can prove that $\|f'_1\|_\infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.