Prove that $f_n$ does not converge uniformly to any function $f\colon(-1,1)\to {\Bbb R}$.

Question

Let $(f_n)_n$ be a sequence of functions $f_n\colon(-1,1)\to \mathbb{R}$, defined by $f_n(x)=x^n$. Prove that $(f_n)_n$ does not converge uniformly to any function $f\colon(-1,1)\to\mathbb{R}$.

How to do with the statement "any function $f\colon(-1,1)\to\mathbb{R}$"? I just do not how to start. Thank you very much!

Answer 1

Since, for each $x\in(-1,1)$, $\lim_{n\to\infty}f_n(x)=0$, the only function $f\colon(-1,1)\longrightarrow\mathbb R$ such that $(f_n)_{n\in\mathbb N}$ could possibly converge uniformly to is the null function. But $(\forall n\in\mathbb{N}):\sup f_n=1$, and therefore your sequence does not converge uniformly to the null function.

Answer 2

Step 1: identify the only possible limit $f$.

Note that $f_n$ converges pointwise on $(-1,1)$ to some $f$ (easy to find). If there is uniform convergence, it has to be to this same $f$. (Why?)

Step 2. Show there isn't uniform convergence to this $f$.

You want to show that $\lVert f_n-f\rVert_\infty = \sup_{x\in(-1,1)}\lvert f_n(x) - f(x)\rvert$ does not converge to zero. To do so, it is enough to find a sequence $(x_n)_n$ such that $\lvert f_n(x_n) - f(x_n)\rvert$ does not converge to zero. (For instance, s.t. $\lvert f_n(x_n) - f(x_n)\rvert$ converges to something else, a positive number).

Spoilers:

For the first step, $f$ is the all-zero function.

and

For the second step, take $x_n = 1-1/n$.