The norm of $(Af)(x)=\int_0^x f(t)\,dt$ 
For $f \in L^1[0,1]$ let $A:L^1[0,1] \rightarrow L^1[0,1]$ be given by
  $$(Af)(x)=\int_0^x f(t)\,dt, \quad0\le x \le 1.$$
  Show that $A$ is a continuous linear operator and calculate its norm.

I showed continuity, equivalently boundedness in this way: 
$$\|Af\|=\left\|\int_0^xf(t)\,dt\right\|=\int_0^1\left| \int_0^xf(t) \, dt \right| \,dx \le\int_0^1\int_0^x|f(t)| \, dt \, dx \le\int_0^1\|f\| \, dx=\|f\| $$ 
hence the operator is bounded with a constant $C=1$ and thus $\|A\|\le1$.
However, I'm struggling with showing that indeed $\|A\|=1$. I was advised to use $$f_n = n 1_{[0,{1 \over n}]}$$ but I have two questions. Firstly, how do I come up with such a function and what do I use it for?

It is clear that $(\forall n\in\mathbb{N}):\|f_n\|=1$. On the other hand, if $n\in\mathbb N$ and $x\in[0,1]$,$$(Af_n)(x)=\int_0^xf_n(t)\,dt=\left\{\begin{array}{l}nx&\text{ if }x\leqslant\frac1n\\1&\text{ otherwise.}\end{array}\right.$$Therefore, $\|Af_n\|=1-\frac n2$. So, you have a sequence of elements of your space with norm $1$ such that the norms of their images tendo to $1$. So, $\|A\|\geqslant1$. Since you already know that $\|A\|\leqslant1$, $\|A\|=1$.
