Continuity of a linear functional on various spaces. Let $X=C[0, 1]$ be the space of all real valued continuous function on $[0, 1].$ Let $T: X\rightarrow \mathbb{R}$ be a linear functional defined by $T(f)=f(1)$.
Let $X_1=(X, \vert \vert .\vert \vert_1)$
Let $X_2=(X, \vert \vert .\vert \vert_2)$
Let $X_3=(X, \vert \vert .\vert \vert_{\infty})$
where $T$ is continuous?
My thoughts:
$\vert T(f) \vert \leq \vert f(1) \vert \leq \vert \vert f\vert \vert_{\infty}$. i.e $T $ is continuous on $X_3$ and as $ \vert \vert f \vert \vert_1 \leq \vert \vert f \vert \vert_{\infty}$, so $T$ is continuous on $X_1$, but what about $X_2?$
Please help.
 A: It seems that you are looking for the norms $\|f\|_1=\int\limits_0^1|f(t)|dt$ and $\|f\|_2=\sqrt{\int\limits_0^1|f(t)|^2dt}$. The functional $T(f)=f(1)$ isn't continuous with respect to these norms because $f_n(t)=t^n$ define continuous functions with $\|f_n\|_j\to 0$ for $j=1,2$ (just calculate the integrals) but $T(f_n)=f_n(1)=1$.
A: The inequality $\|\cdot\|_1 \le \|\cdot\|_\infty$ does not help. From boundedness in the $\|\cdot\|_\infty$ norm you cannot conclude boundednes in the $\|\cdot\|_1$ norm. Indeed, let $\|T\|_\infty$ be the operator norm of $T$ with respect to $\|\cdot\|_\infty$.
$$|Tf| \le \|T\|_\infty \|f\|_\infty \stackrel{?}{\le } \|T\|_\infty \|f\|_1$$
For that you would need the reverse inequality $\|\cdot\|_\infty \le \|\cdot\|_1$ which does not hold in general.
$T$ is in fact not bounded with respect to $\|\cdot\|_1$.
For example, for $n \in \mathbb{N}$ consider $f_n \in C[0,1]$ given as
$$  f(t) =
\begin{cases}
0,  & \text{if $x \in \left[0, \frac{n-1}n\right]$} \\
2n^2t+2n(1-n), & \text{if $x \in \left[\frac{n-1}n, 1\right]$}
\end{cases}$$
We have $\|f_n\|_1 = 1$ but $Tf_n = f_n(1) = 2n$ for all $n \in \mathbb{N}$. So there cannot exist a constant $C > 0$ such that:
$$2n = |Tf_n| \le C\|f_n\|_1 = C, \forall n \in \mathbb{N}$$
The situation is similar with $\|\cdot\|_2$. For $n \in \mathbb{N}$ define $g_n \in C[0,1]$ as
$$  g_n(t) =
\begin{cases}
0,  & \text{if $x \in \left[0, \frac{n-1}n\right]$} \\
\sqrt{2n^2t+2n(1-n)}, & \text{if $x \in \left[\frac{n-1}n, 1\right]$}
\end{cases}$$
We have $\|g_n\|_2 = 1$ but $Tg_n = g_n(1) = \sqrt{2n}$ for all $n \in \mathbb{N}$. So there cannot exist a constant $C > 0$ such that:
$$\sqrt{2n} = |Tg_n| \le C\|g_n\|_2 = C, \forall n \in \mathbb{N}$$
Therefore, we conclude that $T$ is not bounded (i.e. continuous) with respect to $\|\cdot\|_1$ and $\|\cdot\|_2$.
