$X = C[0,1]$. $$x_n(t) =
\begin{cases}
nt, & \text{for $0 \leq t \leq \frac{1}{n}$ } \\
2-nt, & \text{for $\frac{1}{n} \leq t \leq \frac{2}{n}$ } \\
0, & \text{for $\frac{2}{n} \leq t \leq 1$} \\
\end{cases}$$
I know $x_n$ doesn't tend to zero strongly, but I don't know how to show it tends to zero weakly in $L^{\infty}$.
Added: A sequence of $\{x_n\}$ in a normed linear space $X$ is said to converge weakly to $x$ if $$\lim_{n \to \infty}l(x_n) = l(x)$$ for every $l$ in $X'$(the dual of $X$, i.e. the collection of all continuous linear functionals.).
Best Answer
I assume that what you want to prove is weak convergence in $C[0,1]$ endowed with $L^{\infty}$ norm. Using Riesz representation, i.e. the dual of $C[0,1]$ can be identified with bounded Borel measures, we need only to take care of convergence $\int x_n d\mu \to 0$ for any measure $\mu$. But this is equivalent for the sequence $(x_n)$ to being uniformly bounded and pointwise convergent, via dominated convergence theorem.