Uniform and pointwise convergence of $ f_n(x)= \begin{cases} 0,\ x\leq n \\ x-n, \ x\geq n \end{cases} $ and $f_n(x)=x^n-x^{2n}$

Question

I need help with this problem:

For the next sequence {$f_n$}, determine the pointwise limit of {$f_n$} on the interval and indicate if {$f_n$} converges uniformly to that function.

1. $ f_n(x)= \begin{cases} 0,\ x\leq n \\ x-n, \ x\geq n \end{cases} $ on $[a,b]$ and on $\mathbb{R}$
2. $f_n(x)=x^n-x^{2n}$ on $[0,1]$

For the first one I know that when $x\geq n$, {$f_n$} would be the identity function displaced $n$. I don't know how to take the pointwise limit of the function. How do I find the pointwise limit and how do I determine if it converges uniformly?

For the secon one I did the following: $f(x)=\lim_{n\to\infty} x^n-x^{2n} = \lim_{n\to\infty} x^n - \lim_{n\to\infty} x^{2n} = 0$, so the pointwise limit is $0=f(x)$ I don't know how to determine if it converges uniformly, how do I do it? Thanks.

Answer 1

You can prove that convergence to zero is not uniform on $[0,1]$ by finding the point of max of $f_n$ ($x_{max}=1/\sqrt[n]{2}$) and then proving that $f_n(x_{max})$ does not approach zero. $$ |f_n(x_{max})-0|=f_n(2^{-1/n})=1/4. $$

Answer 2

For the first case, $$f_n(x)= \begin{cases} 0,\ x\leq n \\ x-n, \ x\geq n \end{cases}$$

the pointwise limit is $f(x) = 0$. To see this, take $x$, then for large enough $n$, $x \leq n$ and therefore $f_n(x) = 0.$ Now, for uniform convergence, if we go back to the definition, we need to show that: $$ \forall \epsilon > 0, \exists N, \forall n \geq N, \forall x \in [a,b], f_n(x) < \epsilon.$$ This holds for $N \geq b$ (assuming $b >0$ for simplicity). Now, uniform convergence does not hold on $R$. Consider for example $x_n = n+1$, then $f_n(x_n) = 1.$