Explanation for why $f(x) = x - x^n$ does not converge uniformly?

Question

I know intuitively that $f_{n}(x) =x - x^n$ does not converge uniformly.

It converges pointwise to $f(x) = \begin{cases} x, & 0 \leq x \lt 1 \\ 1, & x = 1 \end{cases}$

I know that since is it discontinuous at x = 1, it does not converge uniformly, but I am struggling to provide an explanation as to why.

I am using the following definition for uniform converge: $f_n(x)$ converges uniformly to $f(x)$ iff for all $\epsilon \gt 0$, there exists $N$ s.t. $n \gt N$ => $||f(x) - f_n(x))||_∞\lt \epsilon$.

Answer 1

If $f_n$ converges uniformly the limit has to be $f(x)$. Let $0<\epsilon < \frac 1 e$. Then there exists $n_0$ such that $|f_n(x)-f(x)| <\epsilon$ for $0\leq x <1$ and $n \geq n_0$. This means $|x^{n}|<\epsilon$ for $0\leq x <1$ and $n \geq n_0$. Put $x=1-\frac 1 n$ and let $n \to \infty$ to get a contradiction.

Answer 2

Let's try simpler functions, $f_{n}(x) = x^{n}$. These converge pointwise on $[0, 1]$ to the function $f$ s.t. $f(1) = 1$ and $f(x) = 0$ for $0 \leq x < 1$.

No matter how small we choose a positive $\epsilon > 0$, we will have $$ \lim_{n \rightarrow \infty} f_{n}(1 - \epsilon) = 0. $$ And this explains the absence of uniform convergence on $[0, 1]$.