Let $n \in \mathbb{N}$. Consider the function $$f_n: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto 1 - \frac{3}{n} \left(x-1+\frac{1}{n}\right)^2 + \frac{x}{n}. $$ For all $x \in \mathbb{R}$ we have $\lim_{n \to \infty} f_n(x) = 1.$ I need to determine whether the convergence is uniformly or not.
I think it is not uniformly convergent. I tried to prove this, but I'm having trouble finding the correct $x$ and the correct estimate. We need to prove that $$\exists \epsilon > 0, \forall n_0 \in \mathbb{N}, \exists x \in \mathbb{R}, \exists n \geq n_0: |f_n(x) - 1| \geq \epsilon. $$ I let $\epsilon = 1$. Let $n_0 \in \mathbb{N}$ be arbitrary. For all $x \in \mathbb{R}$ we have that $$ | f_n(x) - 1| = \left| - \frac{3}{n}\left (x-1+\frac 1n\right)^2 + \frac{x}{n} \right|. $$ Now I don't know how to pick $x$ and $n_0$ to get this bigger than $1$. Any help is appreciated.
Uniform convergence is equivalent to $\sup \limits _{x \in \Bbb R} | f_n - 1 | \to 0$. But
$$\begin{align} \sup \limits _{x \in \Bbb R} | f_n - 1 | &=& \\ \frac 1 n \sup _{x \in \Bbb R} \left| 3\left(x-1 + \frac 1 n\right)^2 - x \right| &=& \\ \frac 1 n \sup _{x \in \Bbb R} \left|3x^2 + \left(-6 + \frac 6 n -1\right)x + 3 \left(-1 + \frac 1 n\right)^2 \right| &=& \\ \frac 1 n \infty &=& \\ \infty , \end{align}$$
which shows that, indeed, the convergence is not uniform.