The sequence $f_n=x^n$ is not weakly convergent in $C[0,1]$ Let's consider the sequence $f_n=x^n$ for $n \in \mathbb{N}$ in $C[0,1]$ equipped with the usual supremum norm.
How can we show that $f_n$ does not converge weakly in $C[0,1]$ without using an explicit description of the dual (e.g. as a space of measures)?
This was a question in an exam in functional analysis this semester and I can't really figure out a way how to solve this without knowing how the dual of $C[0,1]$ looks like.
 A: For each $x\in[0,1]$, define the evaluation functional $T_x:C[0,1]\to\mathbb R$ as $T_x(g)\equiv g(x)$ for each $g\in C[0,1]$. It is clear that $T_x$ is linear and it is bounded, since $|T_x(g)|=|g(x)|\leq\|g\|_{\infty}$ for each $g\in C[0,1]$, so the operator norm of $T_x$ is finite (and not greater than $1$). This means that $T_x\in C[0,1]^{\star}$ (the dual space) for all $x\in[0,1]$.
Now suppose that there exists some $f\in C[0,1]$ such that $(f_n)_{n\in\mathbb N}$ converges to $f$ weakly. Then, one has $$T_x(f_n)\to T_x(f)\quad\text{as $n\to\infty$ for each $x\in [0,1]$}.$$
This implies, in particular, that $$T_1(f_n)=f_n(1)=1\to T_1(f)=f(1),$$ so that $f(1)=1$. But if $x\in[0,1)$, one has $$|f(x)|=|T_x(f)|\leq|T_x(f)-T_x(f_n)|+|T_x(f_n)|=|T_x(f)-T_x(f_n)|+|x^n|\to 0,$$ so that $f(x)=0$. Therefore,
\begin{align*}
f(x)=\begin{cases}0&\text{if $x\in[0,1)$,}\\1&\text{if $x=1$,}\end{cases}
\end{align*}
so that $f$ is not continuous. This contradiction reveals that the sequence $(f_n)_{n\in\mathbb N}$ cannot have a weak limit in $C[0,1]$.
