I have a problem:
Suppose that $E$ be a normed space over $\mathbb{R}$ and $E= \{f: [0,1] \to \mathbb{R}\ \text{is continuous and such that}\ f|_{[0, \delta]}=0, \text{with}\ \delta=\delta(f)>0 \}$
with norm: $\| f \|_{\infty} = \sup_{x\in [0,1]} |f(x)|$
For linear functional consequence $f \mapsto \varphi_n(f)=nf(\frac{1}{n}), \forall n \ge 1$
I'm wondering whether $\{\varphi_n\}_{n \ge 1}$ is weakly bounded iff $\{\varphi_n\}_{n \ge 1}$ is uniformly bounded in $E'$ (the dual space of $E$).
A subset $B\subset E'$ is weakly bounded, iff for each $x\in E$ the set of real numbers $$ \{ |S(x)|; S\in B \} $$ is bounded. $B$ is called uniformly bounded (or norm-bounded or strongly bounded), iff the bound is uniform for $||x||\leq 1$, i.e. the set $$ \{ |S(x)|; S\in B, ||x||\leq 1 \} $$ is bounded in $\mathbb{R}$. In your case, for a fixed $f\in E$, the set $$ \{ |\varphi_n(f)|; n\in \mathbb{N}\} $$ is bounded, as only finitely many $\varphi_n(f)$ are non-zero. Now let for $m\in\mathbb{N}$ $g_m\in E$ with $||g_m||\leq 1$ and $g_m(1/m)=1$ (think why such $g_m$ exist). Now conclude why your set in not uniformly bounded.