Consider the sequence ${f_n}$ with $n \ge 2$, defined on $[0,1]$ by Consider the sequence ${f_n}$ with $n \ge 2$, defined on $[0,1]$ by
$$f_n(x) = 
\left\{
\begin{array}{c}
n^2x, &0 \le x \le \frac{1}{n} \\ 
2n - n^2x, &\frac{1}{n} < x \le \frac{2}{n} \\
0, &\frac{2}{n} < x \le 1
\end{array}
\right.$$
a. Sketch the graph of $f_n$ for $n = 2, 3,$ and $4$
b. Prove that $lim_{n\to\infty}f_n(x) = 0$ for each $x \in [0,1]$
c. Show that $\int_0^1f_n(x)dx = 1$ for all $n = 2,3,...$
I'm not sure on how to sketch a graph for a). Would I be right in having 3 sections for $0 \le x \le \frac{1}{n}$, $\frac{1}{n} < x \le \frac{2}{n}$, and $\frac{2}{n} < x \le 1$ and then evaluating $f_n(x)$ for $n = 2, 3,$ and $4$ so that there's 3 graphs within each section? I'm also not sure how to do b) and c) as well. I'm just learning pointwise limits so I'm not familiar with these yet, so any hints/help would be greatly appreciated!
 A: This is the classical counterexample given to show that norms $||.||_1$ and $||.||_{\infty}$ aren't equivalent over $\mathcal{C}[0,1]$, because the family $(f_n)_{n\in \mathbb{N}}\in \mathcal{C}[0,1]$ and $||f_n||_1=1$, but there not exists any $m>0$ such that 
$$m||f_n||_{\infty} \leq ||f_n||_{1}\leq ||f_n||_{\infty}$$  To show that suppose that this constant exists, we would have that $m\leq \frac{1}{n}, \forall n \in \mathbb{N}$, but that's not possible.
a) The graph of the functions $f_n$ is zero in everywhere except an isosceles triangle with basis $[0,2/n]$ and high $n$.
(b) Easy once you did (a). Fix the point $x$ and see what happens with the limit.
(c)$\int_0^{1} f_n(x)=(2/n).(n/2)=1$ (Area of the triangle)
A: a) Note that for the first two pieces only have a single power of $x$, so the graphs of these pieces are straight lines so, in order to graph these parts, it is enough to know where they begin and end.
b) As $n$ gets bigger, what happens to the three pieces? Which one gets bigger? If you fix $x$, can you guarantee that it will eventually be in this piece (i.e. for a big enough value of $n$)?
c) Once you draw the graphs in a), you will realise that the area between the graph and the $x$-axis is easy enough to calculate without the need to do any integration.
