Proving uniform convergence of $f_n(x) = \frac{x^n}{1+x+x^n}$ on various intervals

Question

Prove (or disprove) $\frac{x^n}{1+x+x^n}$ is uniformly continuous on:

a) $[0, \infty]$

b) $[0, \alpha]$ where $0<\alpha < 1$

c) and $[\beta, \infty)$ where $1<\beta < \infty$.

Here is the definition of uniform convergence:

$(f_n)$ converges uniformly on domain $A$ if, for every $\epsilon > 0$, there exists an $N\in\mathbb{N}$ such that $|f_n(x)-f(x)|<\epsilon$ whenever $n\geq N$ and $x\in A$.

My work thus far:

a) I don't believe this converges uniformly on that interval, as different subintervals produce a different function. On $[0, 1)$ is converges to $f(x)=0$, on $[1, 1]$ it coverges to $f(x)=\frac{1}{3}$, and on $(1, \infty)$ it converges to $f(x)=1$. Since it converges to different functions, it cannot possibly converge uniformly.

b) I must prove it converges uniformly to $f(x)=0$.

$|\frac{x^n}{1+x+x^n} - 0| = |\frac{x^n}{1+x+x^n}|$

Not sure where to go from here.

c) I must prove it converges uniformly to $f(x)=1$.

$|\frac{x^n}{1+x+x^n} - 1| = |\frac{x^n-1-x-x^n}{1+x+x^n}| = |\frac{-1-x}{1+x+x^n}| = |\frac{1+x}{1+x+x^n}|\leq |\frac{1+x}{x^n}|$

And I'm not sure where to go from there.

In all cases where I'm stuck, I can't seem to shake that $x$ since $n$ is in the exponent. For uniform convergence, your choice of $N$ cannot depend on $x$, only $\epsilon$. Any help would be appreciated.

Answer 1

a) The argument is not

Since it converges to different functions, it cannot possibly converge uniformly.

You did right to consider the pointwise limit, so $$ f(x)=\begin{cases} 0 & 0\leq x<1\\ \frac13 & x=1\\ 1 & x>1 \end{cases}. $$ Now you have to argue indirectly:

If $(f_n)_n$ converges uniformly, then the pointwise limit function $f$ is continuous by Weierstraß theorem. But $f$ is discontinuous at $1$, so $(f_n)_n$ doesn't converge uniformly.

b) You are almost done. Consider $1+x+x^n>1$ hence $\left|\frac{x^n}{1+x+x^n}\right|\leq x^n\leq \alpha^n$.

c) You are almost done. Consider $x\geq \beta>1$ hence $$ \left|\frac{1+x}{x^n}\right|\leq \frac{x+x}{x^n}=\frac2{x^{n-1}}\leq \frac2{\beta^{n-1}} $$

Answer 2

A good way to approach this problem is to see that most of the time the points to look for when we suspect that uniform convergence does not hold, over $\:\mathbf R_+ ,\:$ are $\:x_n\in\{\frac{1}{n},1\pm\frac{1}{n},n\}.$

Using this procedure has always brought me non-trivial results.