Show that $ T \notin X' $ if $ X = C([0,1]) $ is equipped with the norm $ \| f \|_{L^{2}} \stackrel{\text{df}}{=} \sqrt{\int_{0}^{1} |f|^{2}} $. Let $ T $ be an operator on $ X = C([0,1]) $ defined by $ T(f) \stackrel{\text{df}}{=} f(0) $. I want to show that $ T \notin X' $ (the dual space of $ X $) if $ X $ is equipped with the norm
$$
\| f \|_{L^{2}} \stackrel{\text{df}}{=} \sqrt{\int_{0}^{1} |f|^{2}}.
$$
I know that if $ T $ isn’t bounded, then it can’t be an element of $ X' $, but I couldn’t find an $ f $ so as to make $ T $ unbounded. Can you help me? Thank you for your help.
 A: You need to construct a sequence of functions $(f_n)$ such that $\sup_n\lVert f_n\rVert<\infty$ but $f_n(0)\geq n$ for each $n\in\mathbb N$. Try the functions $f_n$ which take the value $n$ at $x=0$ and decreases to $0$ at $1/n^2$. Then show that $(f_n)$ has the desired properties.
A: Look the sequence of functions $g_n(x) \in C[0, 1]$ defined by
$g_n(x) = 1 - nx, \; \; x \in [0, \dfrac{1}{n}], \tag{1}$
$g_n(x) = 0, \;\; x \in (\dfrac{1}{n}, 1].  \tag{2}$
We observe that
$Tg_n = g_n(0) = 1 \tag{3}$
for every $g_n(x)$; we compute the $L^2$ norm of $g_n(x)$:
$\Vert g_n(x) \Vert_{L^2}^2 = \int_0^1 (g_n(t))^2 dt = \int_0^{1/n} (1 - nt)^2 dt; \tag{4}$
since
$(-\dfrac{1}{3n}(1 - nx)^3)' = (1 - nx)^2, \tag{5}$
we have
$\Vert g_n(x) \Vert_{L^2}^2 = (-\dfrac{1}{3n}(1 - nx)^3 \mid_0^{1/n} = \dfrac{1}{3n}, \tag{6}$
whence
$\Vert g_n(x) \Vert_{L^2} = \dfrac{1}{\sqrt{3n}} \to 0 \;\; \text{as} \; n \to \infty. \tag{7}$
It follows from (3) and (7) that $T$ cannot be bounded in the $L^2$ norm on $C[0, 1]$, for if there were such a bound $C$ then
$1 = \vert 1 \vert = \vert Tg_n \vert \le C\Vert g_n \Vert_{L^2} = \dfrac{C}{\sqrt{3n}} \to 0 \tag{8}$
as $n \to \infty$; the absurdity of (8) precludes the existence of $C$; hence, $T$ cannot be bounded in the $L^2$ norm on $C[0, 1]$; $T \notin X'$.
