The sequence of functions $\{f_n\}_n$ is defined on $[0,1]$ by:
$$f_n(x) = a_n \times (1 - nx),\ {\rm\ if}\ x \in ]0,\frac{1}{n}],$$ and $f_n(x) = 0$ otherwise, where $(a_n)_n$ is a positive sequence of real numbers.
Q: How can we choose the sequence $(a_n)$ so that $\{f_n\}_n$ converges uniformly to $0$ on $[0,1]$?
Attempt:
It's easy to verify that $\{f_n\}$ converges pointwise to $0$ on $[0,1]$. Now let $n \in \mathbb N$ be given and let's study the behaviour of $f_n$. For $x \notin ]0, \frac{1}{n}]$, there's nothing to study. So let $x \in ]0, \frac{1}{n}]$. We have $f'_n(x) = -n \times a_n \le 0$, for all $x$, so $f_n$ is decreasing. And eventually, $$\sup_{x \in ]0, \frac{1}{n}]} |f_n(x)| = \lim_{x \to 0} f_n(x)$$
To have uniform convergence, we should have that quantity convergent to $0$. As $\lim_{x \to 0} f_n(x) = a_n$, we should then have $a_n$ convergent to $0$.
Is there something missing? Does this fully answer the question?
Thanks.
Your solution is correct but it is very specific to this problem. In general what you can do is as follows.
Here is another way to view uniform convergence. Let $\{f_n\}$ be a sequence of functions defined on a set $E$. If $\{f_n\}$ converges uniformly to $0$ on $E$, then given $\epsilon >0$ there exists $N$ such that \begin{align*} |f_n(x)| < \epsilon & & (n\geq N; x \in E). \end{align*} This implies \begin{align*} \sup_{x\in E}|f_n(x)| \leq \epsilon & & (n\geq N). \end{align*} Hence if $\{f_n\}$ converges uniformly to $0$ on $E$, then \begin{align} \lim_\limits{n\to \infty}\sup_{x\in E}|f_n(x)| = 0. & & (1) \end{align} Conversely, it is not difficult to show that if $(1)$ holds then $\{f_n\}$ converges uniformly to $0$ on $E$.
Since you have already noticed that $\{f_n\}$ converges pointwise to the $0$ function. Now by the above observation the convergence is uniform if and only if \begin{align} \lim_\limits{n\to \infty}\sup_{x\in E}|f_n(x)|= \lim_\limits{n\to \infty}a_n = 0 \end{align}