Is the sequence $(f_n)$ convergent? weakly convergent? in $(C[0,1])'$ Is the sequence $(f_n)$ convergent in $(C[0,1])'$? Weakly convergent? 
$$f_n(x)=n \int^{1/n}_0 x(t)dt , n\in \mathbb{N}.$$
Attempt:
First I tried to show: $f_n\rightarrow f(x):=x(0)$
Then, I tried to find $g_n,g\in BV[0,1]:f_n(x)=\int^{1/n}_0 x(t)dg_n(t)$, $f(x)=\int^{1/n}_0 x(t)dg(t)$.
Next, let $h\in(C[0,1])''$ such that $h(g)=g(0+)-g(0)$. Then how to show $h(g_n) \rightarrow h(g)$ does not hold.
 A: Preliminaries. Using mean value theorem we can show that $(f_n)$ weak-$^*$ converges to $\delta_0$. Indeed, fix $x\in C([0,1])$. By mean value theorem there exist $\xi_n\in [1,n^{-1}]$ such that 
$$
\int_0^{n^{-1}} x(t)dt=x(\xi_n)(n^{-1}-0)=n^{-1}x(\xi_n).
$$
Since $\xi_n\in[0,n^{-1}]$, then $\xi_n\to 0$. So we get 
$$
f_n(x)=n\int_0^{n^{-1}} x(t)dt=x(\xi_n)\to x(0)=:\delta_0(x).
$$ 
Since $x\in C([0,1])$ is arbitrary we proved that $(f_n)$ weak-$^*$ converges to $\delta_0$. 
Absence of strong convergence. Assume $(f_n)$ strongly converges to some $\mu\in C([0,1])^*$. Then $(f_n)$ weak-$^*$ converges to $\mu$. From paragraph $1)$ we know that weak-$^*$ limit of $(f_n)$ is $\delta_0$. Hence $\mu=\delta_0$. For each $n\in\mathbb{N}$ consider $x_n(t)=\max(1−|1−2nt|,0)$. Clearly $\Vert x_n\Vert=1$ and what is more 
$$
(f_n-\delta_0)(x_n)=n\int_0^{n^{-1}}(1−|1−2nt|)dt=n\frac{1}{2}n^{-1}\cdot 1=\frac{1}{2}
$$
Hence
$$
\Vert f_n-\delta_0\Vert\geq \frac{|(f_n-\delta_0)(x_n)|}{\Vert x_n\Vert}=\frac{1}{2}
$$
This ineqaulity shows that $\Vert f_n-\delta_0\Vert\not{\to} 0$, so $(f_n)$ doesn't convergence strongly to any $\mu\in C([0,1])^*$
Absence of weak convergence. I will use the following criterion of Grothendieck

The sequence of measures $(\mu_n)\subset C([0,1])^*$ is weakly convergent iff $\mu_n(U_n)\to 0$ for every sequence of pairwise disjoint open sets $(U_n)$

By Riesz-Markov theorem each $f_n$ can be identified with some unique measure $\mu_n$. In fact $\mu_n(A)=n\lambda([0,n^{-1}]\cap A)$ for each Borel set $A$, where $\lambda$ is the Lebesgue measure on $[0,1]$. Indeed, for every $x\in C([0,1])$ we have
$$
\int_0^1 x(t)d\mu_n(t)=\int_0^1 x(t) nd\lambda([0,n^{-1}]\cap -)(t)=n\int_0^{n^{-1}}x(t)dt
$$
Let $(V_n)$ be any sequence of disjoint open subsets of $[2^{-1},1]$. Let $\varphi$ be any bijection between $\mathbb{N}\setminus\{2^k:k\in\mathbb{N}\}$ and $\mathbb{N}$. Define
$$
U_n:=
\begin{cases}
(2^{-k-1},2^{-k}) &\quad\mbox{ if }\quad n=2^k\\
V_{\varphi(n)}    &\quad\mbox{ otherwise }
\end{cases}
$$
Clearly, $(U_n)$ is a sequence of disjoint open subsets of $[0,1]$. Now for subsequence $n_k=2^k$ we have
$$
\mu_{n_k}(U_{n_k})
=n_k\lambda([0,n_k^{-1}]\cap U_{n_k})
=2^k\lambda([0,2^{-k}]\cap (2^{-k-1},2^{-k}))
=2^k\lambda((2^{-k-1},2^{-k}))
=2^{-1}
$$
Therefore we found subsequence of $(\mu_n(U_n))$ which doesn't converge to $0$. By Grothendieck criterion $(\mu_n)$ doesn't converge weakly. Neither does $(f_n)$.
