Equivalent convergent conditions in $C([0,1])$ I am doing problems in my instructor's notes. I am stuck at the following problem.
"Let $f_n\in C([0,1])$ for $n=1,2,\cdots$. Show that the
  following two statements are equivalent:
1) For every $\lambda\in C([0,1])^*$ we have $\lambda(f_n)\to
      0$ as $n\to\infty$,
2) $f_n(x)\to 0$ for every $x\in [0,1]$ and $\sup
        ||f_n||_\infty<\infty$."
$(1) \implies (2):$ It turns out I did this part wrong. I used Uniform Boundedness Principle for $f_n$, which is not linear.
$(2) \implies (1):$ I am stuck at this one. I want to prove that $f_n\to 0$ and the rest would be easy. we have
$$
lim_{n\to \infty} ||f_n|| = lim_{n\to \infty} \sup_{x\in [0,1]} \{|f_n(x)| \}
$$
here I wish I can interchange lim and sup and we are done. Is this approach true and how do we do it?
Thanks in advance
 A: $(1) \implies (2):$ Take $x \in [0,1]$. Consider the operator $\delta_x$ which maps each $f \in C([0,1])$ to its value $f(x)$, i.e. $\delta_x(f) = f(x).$ Now this operator is centainly linear. It is also bounded as
$$|\delta_x (f)| = |f(x)| \leq ||f||_\infty.$$
Hence $\delta_x \in C([0,1])^*$. From $(1)$ we get
$$f_n(x) = \delta_x (f_n) \rightarrow 0.$$
$f_n$ are unifromly bounded by the uniform boundedness principle as for each $\lambda \in C([0,1])^*$ the sequence of scalars $(\lambda(f_n))_{n=1}^\infty$ is bounded (convergent sequences of scalars are always bounded), and hence $(f_n)_n$ is weakly bounded.
$(2) \implies (1):$ Take $\lambda \in C([0,1])^*$. Then $\lambda$ is a complex (or signed) Radon measure by the Riesz representation theorem. By $(2)$ we have that $f_n \rightarrow 0$ a.e. and that $\sup_n ||f_n|| := S < \infty$. But then we can use Lebesgue dominated convergence theorem with a constant funcion equal to $S$ as a majorant.
$$\lambda(f_n) = \int_0^1 f_n(x) d \lambda(x) \rightarrow 0.$$
