Determine if $f_n(x)=\frac{1-x}{1+x^n}$ converges uniformly on $[0,2]$

Question

I want to determine the uniform convergence of the family of functions $f_n:[0,2]\to\mathbb{R}$ defined by $$ f_n(x)=\frac{1-x}{1+x^n}. $$

My attempt

• Limit of $f_n$

Let $f$ be the limit of $f_n$, then $$ f(x)=\begin{cases} 1-x,& 0\le x <1\\ 0,&1\le x\le 2 \end{cases}. $$ $f$ is continuous on $[0,1]$, but not differentiable at $x=1$.

• Inequalities about $|f_n(x)-f(x)|$

We know that $$ |f_n(x)-f(x)| = \begin{cases} \frac{(1-x)x^n}{1+x^n},&0\le x<1\\ \frac{x-1}{1+x^n},&1\le x\le 2 \end{cases}. $$ If $0\le x<1$, $$ |f_n(x)-f(x)|=\frac{(1-x)x^n}{1+x^n} \le (1-x)x^n, $$ and $(1-x)x^n$ has the maximum $\frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1}$ at $x=\frac{n}{n+1}$. Also, if $1\le x\le 2$, then $$ |f_n(x)-f(x)|=\frac{x-1}{1+x^n}\le \frac{x-1}{x^n} $$ and $\frac{x-1}{x^n}$ has the maximum $\frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}$ at $x=\frac{n}{n-1}$. Thus we obtain \begin{align} |f_n(x)-f(x)| &\le \max\left\{\frac{1}{n}\left(1-\frac{1}{n+1}\right)^{n+1},\frac{n-2}{(n-1)^2}\left(1-\frac{1}{n-1}\right)^{n-1}\right\}\\ &\le \frac{1}{n-1} \end{align} for $n\ge 2$.

• Conclusion

For $\epsilon > 0$, there is $N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon$ by Archimedian property. Then, for any $n\in\mathbb{N}\setminus\{1\}$ and $x\in [0,2]$, if $n > N$, then $$ |f_n(x)-f(x)|\le \frac{1}{n-1}\le \frac{1}{N} <\epsilon. $$ Therefore $f_n$ uniformly converges on $[0,2]$.

"Solution" $f_n$ does not converge uniformly, since $f_n$ is differentiable but $f$ is not differentiable at $x=1$.

However, I think the solution is wrong. For example, $f_n(x)=\frac{\sin nx}{\sqrt{n}}$ converges uniformly on $\mathbb{R}$, whereas $f_n'(x)=\sqrt{n}\cos nx$ does not pointwisely converge. According to Sequence of differentiable functions, there is a sequence of differentiable functions which uniformly converges to a non-differentiable function.

Question Is my attempt correct? If so, is there an easier method to solve the problem?

Answer 1

Your argument is correct. The sequence converges uniformly. The "solution" is wrong. It is not true in general that a uniformly convergent sequence of differentiable functions converges to a differentiable function. An additional assumption is needed. For example, if we know that at some point $x_0$ the sequence of derivatives $f_n'(x_0)$ tends to $f'(x_0)$, then it is true. But not in general.

Answer 2

If the $f_n$ are differentiable and converge uniformly to $f$, then $f$ need not be differentiable (see Wikipedia).

Your pointwise limit is correct! You also correctly bounded $\|f_n-f\|_\infty$ for $0\le x\le 1$ and for $1\le x\le 2$. Thus, your conclusion that the the convergence is uniform holds.

Answer 3

Another method (using very elementary inequalities): let $\epsilon >0$. Then, for $ 1-\epsilon <x\leq 1$ we have $|f_n(x)-(1-x)| \leq |1-x| \leq \epsilon$. For $x \leq 1-\epsilon $ we have $|f_n(x)-(1-x)| \leq \frac {x^{n}}{1+x^{n}} \leq (1-\epsilon)^{n} \to 0$. For $1 \leq x \leq 1+\epsilon$ we get $|f_n(x)|\leq \epsilon$. I will leave the case $1+\epsilon \leq x \leq 2$ to you. [Use the fact that $\frac 1 {(1+\epsilon)^{n}} \to 0$.