Show that $f_n$ does not converge to $f$ in the uniform norm but does converge on the norm $||f||=\sqrt{\int_a^b|f(x)|^2}dx$

Question

Bounded on $f_n:[0,1]\rightarrow\mathbb{R},$

$$ f_n(x)= \begin{cases} 1-nx&\text{if}\, 0\leq x \leq \frac{1}{n}\\ 1&\text{otherwise} \end{cases} $$

I think for the first part I need to show that the $\lim_{n\rightarrow \infty}$ of the supremum of $f_n-f$ does not equal 0 since that is a condition of uniform convergence on the uniform norm.

I am given that $f(x)=1$ pointwise as $n\rightarrow \infty$, so I tried to find the supremum and got

$$ ||f_n(x)-f||=\sup \begin{cases} nx&\text{if}\, 0\leq x \leq \frac{1}{n} \end{cases}\} =1 $$

Is this correct? How do i go from here to the next part?