Let $X=L^2(0,1)$ and let $Tf(x) = f(0)$ with
$D_T= C^1[0,1] $.
We show that $T$ cannot be closed.
Indeed, let $f_n(x) = (1-x)^n$. Then $f_n\in D_T$, $\|f_n\|_{L^2(0,1)}\to 0$,
$t\to \infty$, but $Tf_n = 1$.
Note that in this case $G_T = \{(f, f(0)):\, f\in D_T\}$ and
$\|(f,Tf)\| = \|f\|_{L^2} + |f(0)|$.
Example 2.
Let $X=L^p(0,1)$, $p>1$, $Tf(x) = f'(x)$ and
$D_T = C^1[0,1]$.
We show that $T$ can be closed.
The operator $T$ is not closed as it is defined on $C^1[0,1]$ which is not a closed set
in $L^p(0,1)$. In order to show that $T$ can be closed
if is enough to show that
$$
f_n\in D_T, \quad f_n\to 0, \quad Tf_n \to g \quad \Longrightarrow \quad g=0.
$$
Let $\|f_n\|_{L^p}\to 0$, $f_n\in C^1[0,1]$ and let
$$
\int_0^1 |f_n'(x) - g(x)|^p dx \to 0.
$$
If $1/p + 1/q = 1$, then for any $\varphi\in C_0^\infty(0,1)$ we have
$$
\Big| \int_0^1 (f_n'(x)-g(x)) \varphi(x) \, dx \Big|
\le \Big(\int_0^1 |f_n'(x) - g(x)|^p\,dx\Big)^{1/p}
\Big(\int_0^1 \varphi(x)|^q\,dx\Big)^{1/q} \to 0,
$$
which implies
$$
\int_0^1 f_n'(x) \varphi(x) \, dx \to \int_0^1 g(x) \varphi(x) \, dx.
$$
On the other hand
$$
\Big |\int_0^1 f_n'(x) \varphi(x) \, dx \Big |
\le \Big(\int_0^1 |f_n'(x))|^p\,dx\Big)^{1/p}
\Big(\int_0^1 \varphi(x)|^q\,dx\Big)^{1/q} \to 0, \qquad n\to\infty.
$$
Thus
$$
\Big| \int_0^1 g(x) \varphi(x) \, dx \Big| = 0
$$
for any $\varphi\in C_0^\infty(0,1)$ which means that $g=0$ as
an $L^p$ function.
Definition.
Let $X$ be a linear vector space and let $p:\,X \to {\Bbb R} ({\Bbb C})$
be a linear operator. Then $p$ is called a real(complex) linear functional.
Examples:
a. $p:\, f(x) \to f(0)$, $f\in C(-1,1)$.
b.
Let $f\in L^(0,2\pi)$ and let
$$
f(x) = \sum_{-\infty}^\infty a_n e^{-inx}/\sqrt{2\pi}
$$
be its Fourier series.
Then $p\to a_n$ is a complex linear functional on $L^2(0,2\pi)$.
c.
Let $g\in L^2(0,1)$. We define $p=p_g$ such that
$$
p_g(f) = \int_0^1 f(x)g(x)\, dx.
$$
Definition.
A partially ordered set $S$ is a non-empty set with a relation $"\le"$
satisfying the properties:
(i) $x\le x$
(ii) $x\le y$, $y\le z$ $\Longrightarrow$ $x\le z$.
If for any $x,y\in S$ either $x\le y$ or $y\le x$ then $S$ is called totally ordered.
Definition.
Let $T\subset S$, and $S$ be partially ordered. The element
$x\in S$ is called an upper bound if for any $y\in S$, we have $y\le x$.
Definition.
Let $S$ be partially ordered. The element
$x\in S$ is called maximal if for any $y\in S$ the relation $x\le y$
implies $y\le x$.
Theorem.
(Zorn's lemma)
If $S$ is a partially ordered set s.t. every totally ordered subset
has an upper bound, then $S$ has a maximal element.
Theorem.
(AFr Th 4.8.1.) (Hahn-Banach lemma)
Let $X$ be a real linear vector space space and let $p$ be a real functional
(not necessary linear) on $X$ s.t.
$$
p(x+y) \le p(x) + p(y) \qquad p(\lambda x) = \lambda p(x), \qquad \lambda>0, \quad
x,y\in X.
$$
Let $f$ be a real linear functional on a linear subspace $Y\subset X$ s.t.
$$
f(x) \le p(x), \qquad \forall x\in Y.
$$
Then there exists a real linear functional $F$ on $X$ s.t.
$$
F(x)= f(x), \qquad x \in X, \qquad {\rm and } \qquad F(x)\le p(x),
\quad \forall x\in X.
$$
Home exercises.
1. (ex. 4.7.6 from AFr)
Prove that $d/dx$ cannot be extended into a linear operator from
$L^p(0,1)$ into itself.
2. (ex. 4.7.7 from AFr)
Let $C^m(\Omega)$ be a normed space equipped with
$$
\|u\|_m = \sup_{\Omega} \sum_{0\le |\alpha|\le m} |D^\alpha u|,
$$
where $\Omega\subset{\Bbb R}^n$. Define $C^*(\Omega)$ the subset
of $C^\infty(\Omega)$ of all functions $u$ s.t. $\|u\|_m<\infty$ $\forall$ $m\ge 1$.
Introduce in $C^*(\Omega)$ a metric $\rho(u,v)=\rho(u-v,0)$, where
$$
\rho(u,0) = \sum_{m=1}^\infty \frac{1}{2^m}\, \frac{\|u\|_m}{1+\|u\|_m}.
$$
Determine if $d/dx: \, C^*(\Omega)\to C^*(\Omega)$ is continuous.
2.
Let $X= L^2(0,2\pi)$, $\varphi_n=\{e^{inx}/\sqrt{2\pi}\}_{n=-\infty}^\infty$. Let
$\psi\in X$ be an element which is not a finite linear combination
of $\varphi_n$ and let $D$ be a set of all finite linear conbinations
of $\{\varphi_n\}$. Denote by $T$ the operator
$$
T\Big(b\psi + \sum_{i=-M}^N c_j\varphi_j\Big) = b\psi.
$$
Show that ${\overline G_T}$ is not a gragh.
(Namely, show that
$(\psi,\psi)\in {\overline G_T}$ and $(\psi,0)\in {\overline G_T}$.)