I'd like a check for the following exercise I found online
Discuss the $L^{p}(0,1)$, $p\in[1,+\infty]$ convergence of the sequence of functions $$ u_n(x)= \left\{ \begin{array}{c} n^{4/3}x, \quad 0 < x <\frac2n \\ 0, \quad \frac2n \leq x <1 \\ \end{array} \right. $$
Solution
The pointwise limit is $0$:
$\lim_{n \rightarrow +\infty} u_{n}(x)=0$, where $u_n(x)= \mathbb{1}_{(0,\frac2n)}(x) n^{\frac43} x$
In fact, $\forall \epsilon>0 \exists n_0 \text{ s.t } \forall n\geq n_0:|\frac2 n| < \epsilon $
I'll study separately the cases for $p=+\infty$, and $p\geq 1$.
- $p=+\infty$
$\lim_n \sup_{x \in (0,1)} |u_n(x)|=\lim_n \sup_{x \in (0,\frac2n)} n^{\frac43}x=\lim_n 2 n^{\frac13} \rightarrow +\infty$ and thus there's no $L^{\infty}$ convergence in $(0,1)$
- $p \geq 1$
The integral to be computed is $\int_0^1 |u_n(x)|^pdx=\frac{2^{p+1}}{p+1} \cdot n^{\frac{p-3}{3} }$ and taking the limit for $n \rightarrow +\infty$ this is $0$ iff $\frac{p-3}{3}<1$, thus we have $L^{p}(0,1)$ convergence for $p \in [1,6)$
$\square$
You have made two mistakes. $\int u_n^{p}= cn^{4p/3} \frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $\frac {4p} 3 <p+1$ and this is true iff $p < 3$.