Find limit of sequence of functions

Let $\displaystyle f_n\in C[0,1]$ be defined by

$\displaystyle f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}$

Let $\displaystyle \parallel f \parallel _1 = \int_{0}^{1}\mid f(x)\mid\, dx$ and $\displaystyle \parallel f \parallel _2 = \left (\int_{0}^{1}\mid f(x)\mid ^2\, dx \right )^\frac{1}{2}$

Show that $\displaystyle f_n \rightarrow 0$ with respect to $\displaystyle \parallel \cdot \parallel _1$, but $\displaystyle f_n$ is not convergent with respect to $\displaystyle \parallel \cdot \parallel _2$

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Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that $\displaystyle f_n \rightarrow 0$ in both cases.

Choose $\displaystyle x \in [0,\tfrac{1}{n}]$ for some small n.

$\displaystyle \exists \, n_0 > n$ st $\displaystyle x \notin [0,\tfrac{1}{n} ]$ but rather $\displaystyle x \in (\tfrac{1}{n} ,\tfrac{2}{n} )$

$\displaystyle \exists \, \tilde{n}_0 > n_0 $ st $\displaystyle x \notin (\tfrac{1}{n},\tfrac{2}{n} )$ but rather $\displaystyle x \in [\tfrac{2}{n} ,1 )$

$\displaystyle \Rightarrow \, \exists \, \hat{n}_0$ st $\displaystyle f_n = 0 \, \forall \, n \ge \hat{n}_0$

So, clearly, we have that $\displaystyle \forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N}$ st $\displaystyle \parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0$

So what have I missed that stops the above from working, and what is a better way of doing it?