Let $X=(0,1)$ and $u_v(x) = v^{1/p}1_{(0,1/v)}$, where $1 < p < \infty$. It needs to be shown that $\lim_{v \to \infty} \int_X u_v \phi = 0$ for all $\phi \in \mathcal{L}^{p^\prime}$ (The definition of Weak convergence).
As the step functions are dense in $\mathcal{L}^{p^\prime}$ we have step functions $\phi_n$ such that $\|\phi_n - \phi\|_{\mathcal{L}^{p^\prime}} \to 0$. This gives that $\int_X u_v \phi_n \to \int_X u_v \phi$ from Holders inequality and the fact that $\|u_v\|_{\mathcal{L}^{p}}=1$.
It can be shown that $\lim_{v \to \infty} \int_X u_v \phi_n = 0$ for all $n$. The step functions are chosen so that $0=a_0 < a_1 < \cdots < a_n = 1$ and $\phi_n$ takes some constant value, say $\alpha_{n,i}$ on the interval $(a_{i-1},a_i)$. This ensures that $\lim_{v \to \infty} \int_X u_v \phi_n = 0$.
To show that $\lim_{v \to \infty} \int_X u_v \phi = 0$ for all $\phi \in \mathcal{L}^{p^\prime}$ we need an interchange of limits i.e., $\lim_{v \to \infty} \int_X u_v \phi = \lim_{v \to \infty} \lim_{n \to \infty} \int_X u_v \phi_n = \lim_{n \to \infty} \lim_{v \to \infty} \int_X u_v \phi_n$.
To show that limits interchange, it is enough to show that for some large $v$, we have $\int_X |u_v \phi| < \epsilon$. However, this cannot be shown from the Holders inequality.
This is Example 1.17 in the book Calculus of Variations by Bernard Dacorogna. Any help is appreciated. Thanks.
Hint: By Hölder's inequality,
$$\begin{align*} \int_X |u_v \phi| \, dx &= \int_X |u_v| \cdot |1_{(0,1/v)}\phi| \, dx \leq \|u_v\|_p \cdot \|\phi 1_{(0,1/v)}\|_{p'}.\end{align*}$$
Now use monotone convergence in order to deduce that
$$\int_X |u_v \phi| \, dx < \varepsilon$$
for $v$ sufficiently large.