Let: $f_n(x) = \begin{cases} 1 & -n \le x \le n \\ 0 & \text{ otherwise} \end{cases}$ Prove that $\|f_n\|_∞ = 1$, but $\|f_n\|_2 \to \infty$ as $n \to \infty$ and that there is no constant $C$ such that $\|f\|_2 \le C\|f\|_\infty$ for all functions $f$.
My attempt:
I know that for $\|f_n\|_2$ it goes to $\infty$ since the integral will equal to $x$ from the limits $-n \to n$ and if $n = \infty$ we have that $\|f_n\|_2 = \infty$. But why does $\|f_n\|_\infty = 1$? Also, since they both don't approach the same limit it means there is no constant $C$?