TITLE: Convergence of a sequence in $(C[0,1])^*$
QUESTION [6 upvotes]: I would like to determine if the following sequence of linear functionals on $C[0,1]$ converges weakly (i.e. pointwise) and strongly (with respect to the norm) $$ \phi_n(f)=\int\limits_{0}^{1}(nt^2-[nt^2])f(t)dt,$$ where $[x]$ means greatest integer that is less than or equal to $x$. I do not know how to investigate this question. Thanks in advance for your help.

REPLY [4 votes]: The function $h(t)=\{t\}=t-[t]$ is $1$-periodic. The change of variable $u=t^2$ gives
$$\int^1_0f(t)\{nt^2\}\,dt=\frac{1}{2}\int^1_0 \frac{f(\sqrt{u})}{\sqrt{u}}\,h(nu)\,du$$
Notice that  $g(u)=\mathbb{1}_{[0,1]}(u)\frac{f(\sqrt{u})}{\sqrt{u}}$ is integrable for any $f\in\mathcal{C}([0,1])$.  By Féjer's formula,
$$\begin{align}
\frac12\int^1_0 \frac{f(\sqrt{u})}{\sqrt{u}}h(nu)\,du&= \frac12\int_\mathbb{R}g(u)\,h(n\,u)\,du\xrightarrow{n\rightarrow\infty}\frac12\Big(\int^1_0 h(u)\,du\Big)\int_\mathbb{R}g(u)\,du\\
&=\frac14\int^1_0 \frac{f(\sqrt{u})}{\sqrt{u}}\,du=\frac{1}{2}\int^1_0f
\end{align}$$
Therefore, the sequence of measures $\phi_n(dt)=\{nt^2\}\,dt$ converge to $\frac{1}{2}dx$ in $(\mathcal{C}[0,1])^*$ with the weak-* topology.