I need help with this problem: Consider the sequences of functions $(f_n)$ defined by: $$f_n (x)=x^{2} - \frac{x}{n}$$.
- Show that $(f_n)$ converges uniformly to $f$ on$[0, 1]$.
I've already found the pointwise limit, which is $x^{2}$. How do I prove that it converges uniformly?
I tried by saying that for all $\epsilon >0$ there exists an natural number $N$ such that $|f_n (x)-f(x)|<\epsilon$, sí I ended up with $|x^{2} - \frac{x}{n} - x^{2}| = |\frac{x}{n}|$, I think that it is less that $\frac{1}{n}$ because $x$ is in the interval $[0, 1]$, $\frac{x}{n} < \epsilon$, so it converges uniformly to $f$.
$$\sup_{ x \in [0,1]} \mid f(x) - f_n(x) \mid \leq \frac{1}{n} \to 0$$
Thus the sequence converges uniformly