Prove that the series of functions $x_n(t)=n^{\frac{1}{p}}t^n, t\in (0,1)$ converge weakly to $0$ in $L^{p}(0,1),p>1$ So for a series $x_n$to converge weakly in a space $X$ to some value $y$.
Then $\forall x^{*} \in X^{*}$ 
$$x^{*}(x_n)\to^{n \to \infty} x^{*}(x)$$
Okay: So I know that $(L^p[0,1])^* \cong L^{q}[0,1]; \ \ \frac{1}{p}+\frac{1}{q} = 1.$ Therefore:
$$\exists! y(t) \in L^{q}[0,1] \text{ such that:} \ \ x^*(x_n(t)) = \int_{0}^{1}x_n(t)y(t)dt$$
I obvoiusly see from here that if $y(t)$, the series convergenges to zero. But how to I prove that $y(t)$ has to be $0?$
 A: I assume that $1\lt p\lt \infty$. A computation of  the $\mathbb L^p$-norm of $x_n$ show that this quantity can be bounded independently on $n$. Therefore, it suffices to fund $x\in\mathbb L^p$ such that for all $y\in D$, 
$$\lim_{n\to +\infty}\int_0^1x_n(t)y(t)\mathrm dt=\int_0^1x(t)y(t)\mathrm dt,$$
where $D$ is a dense subset of $\mathbb L^{p'}$, $1/p+1/p'=1$. An example for $D$ for which the computations are easily doable is the set of polynomials. In this way, it suffices to compute $\lim_{n\to +\infty}\int_0^1x_n(t)t^k\mathrm dt$ for any fixed $k$ to guess what $x$ is, and show the weak convergence.
A: Now that you know how the elements of $L^p(0,1)^*$ are you must show that:
\begin{equation}
\underset{n \rightarrow \infty}{\text{lim}}{\int^1_0 x_n(t)y(t)dt}=0
\end{equation}
But, using Holder inequality we have:
\begin{equation}
\left|{\int^1_0 x_n(t)y(t)dt}\right|\leq \int_0^1|x_n(t)y(t)|dt \leq \|x_n\|_p\|y\|_q
\end{equation}
Now, suppose that $y\neq0$. So, to conclude, we must show that when $n \uparrow \infty$,  $\|x_n\|_p \rightarrow 0$. But this is simple, because:
\begin{equation}
\|x_n\|_p = \left(\int_0^1nt^{np}dt\right)^{\frac{1}{p}}=\left(\frac{t^{np+1}}{p+1}\right)^{\frac{1}{p}}
\end{equation}
This clearly goes to zero because $t \in (0,1)$!
