As the title says, I am trying to show the weak convergence of the function sequence $ n^{\frac{1}{p}} \chi_{[0,1]}(n x) $, which is supposed to converge to $0$ for $p>1$ but not for $p = 1$. I have found an answer to this question (which was not very helpful to me) under the following post: $f_n(x) = n^{1/p}\chi(nx)$ in $L^p(\mathbb{R})$ weakly converges to 0 for $1 \leq p <\infty$
I understand the use of Holder's inequality but I don't see why that leads to the limit of the weak convergence condition being zero, since we have an inequality, not an equation. Second I can't seem to immediately see why the integral converges at all, even after susbtituting $q$ for $p$ using the equation $ \frac{1}{q}+\frac{1}{p}=1$. Any help would be greatly appreciated.
For $1<p<\infty$, Kavi's solution is very clear. To recap, letting $f_n(x)=n^{1/p}\mathbb{1}_{[0,1]}(nx)=n^{1/p}\mathbb{1}_{[0,1/n]}(x)$, he makes use of Hölder's inequality to get $$\big|\int f_n g\big|\leq\|f_n\|_p\|g\|_q=n^{1/p}\frac1n \|g\|_q=\frac{1}{n^{1-\tfrac1p}}\|g\|_q\xrightarrow{n\rightarrow\infty}0$$ for all $g\in L_q$, where $\tfrac1p+\tfrac1q=1$. This means that $f_n$ converges to $0$ in the weak topology $\sigma(L_p,L_q)$.
For $p=1$, notice that for any continuous bounded function $g$ \begin{align} n\int\mathbb{1}_{[0,1]}(nx) g(x)\,dx&=\int^1_0 g(x/n)\,dx\xrightarrow{n\rightarrow\infty}g(0) \end{align} by dominated convergence. This shows that $f_n$ does not converge to $0$ in the weak topology $\sigma(L_1,L_\infty)$. In fact, no subsequence of $f_n$ converges in $\sigma(L_1,L_\infty)$ for there is not $f\in L_1$ such that $\int f g=g(0)$ for all $g\in C_b(\mathbb{R})$ (The Reisz-Markov representation for example).
In the sense of measures, i.e. $\mu_n(dx)=n\mathbb{1}_{[0,1]}(nx)\,dx$, the sequence $\mu_n$ converges weakly to the probability measure $\delta_0$, which is not absolutely continuous with respect to Lebesgue's measure.