$f_n(x) = \frac{x+x^2}{n}$ is not uniformly convergent

Question

Let $f_n(x) = \frac{x+x^2}{n}$

Then for any real number $a$, $\lim_{n \to \infty} f_n(a) = 0$

Now for seeing the supremum of $\{|f_n(x) -0|: x \in \Bbb{R}\}$

I let $y = \frac{x+x^2}{n}$

Then $y' = \frac{1+2x}{n}$

Then $y' = 0$ only at $x = \frac{-1}{2}$

This implies that maximum can occur only at $\frac{-1}{2}$.

Then the value of $|f_n(x|$ comes out to be $\frac{1}{4n}$ which tends to $0$ as $n$ tends to $\infty$.

That means that $(f_n)$ is uniformly convergent to $0$.

I don't know what's wrong with my working. I saw that it is not uniformly convergent somewhere.

Please point out the mistake.

Answer 1

You have $f_n'(x)=0\iff x=-\frac12$. It turns out that $-\frac12$ is where $f_n$ attains its minimum. It turns out that each $f_n$ is unbounded, since $\lim_{x\to\pm\infty}f_n(x)=\infty$. Therefore, the convergence cannot be uniform: if a sequence $(\varphi_n)_{n\in\Bbb N}$ converges uniformly to a bounded function, then, if $n$ is large enoguh, $\varphi_n$ is bounded.

Answer 2

Just note that $|f_n(n)-0| = \dfrac{n+n^2}{n} \ge n\implies \text{Sup}_{x \in \mathbb{R}}|f_n(x)-0| = +\infty$ .