For a $f(x)=x^n, x\in[0,0.5]$ I know that
$d_\infty (x^n,0)=sup|x^n-0|=\frac{1}{x^n} \rightarrow 0 $, when $n\rightarrow \infty$ and $x\in[0,0.5]$.
For $\{x^n-x^{n+1}\}, x\in[0,1]$ this isn't the case. My teacher goes on to derive
$f_n(x)=x^n-x^{n+1}$ to conclude that $f_n'(x) \geq 0, x\forall \in[0,1]$
Why does she derive it?
thank you!
If the derivative is positive, the function is an increasing function. Since we know that it converges to $0$ on $[0,1]$ and we know that $f_n(0)=0$ for all $n$, then because the function is increasing, the maximum value of $x^n - x^{n+1}$ occurs at $x = 1$. So this can be used to prove uniform convergence.
Alternatively, there is a theorem that you might not have gotten to that states that if a sequence of continuous functions converges to a continuous function on a compact set ($[0,1]$ is compact), then the convergence must be uniform.