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Let us define $h_{n}(t)=\frac{n}{2},t\in [0,\frac{1}{n}]$ and $h_{n}(t)=-\frac{n}{2},t\in [-\frac{1}{n},0]$ and $h_{n}(t)=0$ otherwise. Then $\int f\cdot h_{n}\rightarrow 0$ for each $f\in C[-1,1]$. This means that $(h_{n})_{n}$ is $weak^{*}$-null. Define $g(t)=-1,t\in [-1,0]$ and $g(t)=1,t\in [0,1]$. Then $\int g\cdot h_{n}=1$ for all $n$. This implies that $(h_{n})_{n}$ is not weakly null.