- how to show the subspace of $L^p$ is complete?
Via Cauchy sequences: (i) take a Cauchy sequence in the subspace; (ii) prove that this subsequence converges to some $u\in L^p$; (iii) prove that $u$ indeed belongs to the subspace. This is the basic procedure. However, there are other ways to prove completeness.
- What is the difference of showing a subspace is complete and closed?
To prove that a subspace is closed you have to prove that if a sequence in the subspace converges to $u$, then $u$ is also in the subspace. This is different from showing that every Cauchy sequence is convergent. However, in the context of complete spaces (in particular, in the context of $L^p$ spaces) the two concepts (closed subspace and complete subspace) are the same, that is to say, a subspace is closed if and only if it is complete (see Theorem 2.3-1 in Kreyszig).
- In what sense $(D(T),\|\cdot\|_T)$ is complete?
In the usual sense: if $\{u_n\}$ is a sequence in $D(T)$ which is Cauchy with respect to the norm $\|\cdot\|_T$, then there exists $u\in D(T)$ such that $\|u_n-u\|_T\to 0$.
- I read somewhere while ago (where I really could not remember which book is it) and it says $(D(T),\|\cdot\|_T)$ is complete if and only if $T$ is closed. Intuitively, why?
I don't have an intuitive explanation.
Here there is a sketch.
Addendum
Proposition: Let $(X,\|\cdot\|_X)$ be a Banach space and $T:D(T)\subset X\to X$ a linear operator. Then $(D(T),\|\cdot\|_T)$ is complete if and only if $T$ is closed.
Proof:
$(\Rightarrow)$ Assume that $(D(T),\|\cdot\|_T)$ is complete. Take a sequence $\{u_n\}$ in $D(T)$ such that
$$\|u_n-u\|_X\to0\qquad\text{and}\qquad\|Tu_n-v\|_X\to0$$
for some $u\in X$ and some $v\in X$. As $\{u_n\}$ and $\{Tu_n\}$ are convergent in $X$, we conclude that $\{u_n\}$ is Cauchy in $(D(T),\|\cdot\|_T)$. As $(D(T),\|\cdot\|_T)$ is complete, it follows that there is $x\in D(T)$ such that $\{u_n\}$ converges to $x$ in $(D(T),\|\cdot\|_T)$. As a consequence,
$$\|u_n-x\|_X\to0\qquad\text{and}\qquad\|Tu_n-Tx\|_X\to0.$$
By the uniqueness of the limit, we conclude that $x=u$ and $Tx=v$. Thus, $u\in D(T)$ and $v=Tu$ which implies that $T$ is closed (by Theorem 4.13-3 in Kreyszig).
$(\Leftarrow)$ Assume that $T$ is closed. Take a Cauchy sequence $\{u_n\}$ in $(D(T),\|\cdot\|_T)$. Then $\{u_n\}$ and $\{Tu_n\}$ are Cauchy in $(X,\|\cdot\|_X)$ because
$$\|u_n-u_m\|_X\leq \|u_n-u_m\|_T,\qquad\forall\ n,m\in\mathbb{N}$$
and
$$\|Tu_n-Tu_m\|_X=\|T(u_n-u_m)\|_X\leq \|u_n-u_m\|_T,\qquad\forall\ n,m\in\mathbb{N}.$$
As $(X,\|\cdot\|_X)$ is complete, there are $u,v\in X$ such that
$$\|u_n-u\|_X\to 0\qquad\text{and}\qquad \|Tu_n-v\|_X\to 0.$$
As $T$ is closed, it follows that $u\in D(T)$ and $Tu=v$ (here we are using Theorem 4.13-3 in Kreyszig again). Therefore
$$\|u_n-u\|_T=\|u_n-u\|_X+\|Tu_n-Tu\|_X=\|u_n-u\|_X+\|Tu_n-v\|_X\to 0+0=0.$$
In other words, $\{u_n\}$ converges to $u$ in $(D(T),\|\cdot\|_T)$ and thus $(D(T),\|\cdot\|_T)$ is complete. $\blacksquare$
Addendum 2
In the proof above we have used the fact below (Theorem 4.13-3 in Kreyszig).
$T$ is closed if and only if it has the following property: If $x_n\to x$, where $x_n\in D(T)$, and $Tx_n\to y$, then $x\in D(T)$ and $Tx = y$.
In both parts, $(\Rightarrow)$ and $(\Leftarrow)$, we have passed by a Cauchy sequence in $(D(T),\|\cdot\|_T)$. However, there is no explicit mention to any Cauchy sequence in $X\times Y$ and thus the fact
$\{x_n\}$ is Cauchy in $D(T)$ if and only if $\{(x_n,Tx_n)\}$ is Cauchy in $X\times Y$. $(*)$
is lost. To use it, we should use the definition of closed operator:
$T$ is closed if its graph $\mathcal{G}(T)$ is a closed set in the space $X\times Y$ (equipped with the norm $\|(x,y)\|_{X\times Y}=\|x\|_X+\|y\|_Y$).
From this point of view, we can rewrite the proof as follows.
Proof 2 (expanded pdf's proof):
$(\Rightarrow)$ Assume that $D(T)$ is complete. Take a Cauchy sequence $\{(u_n,Tu_n)\}$ in $\mathcal{G}(T)$. It follows from $(*)$ that $\{u_n\}$ is Cauchy in $D(T)$. As $D(T)$ is complete, it follows that there is $x\in D(T)$ such that $\{u_n\}$ converges to $x$ in $D(T)$. As a consequence,
$$\|(u_n,Tu_n)-(x,Tx)\|_{X\times Y}\to0.$$
which implies that $\mathcal{G}(T)$ is complete (because $(x,Tx)\in\mathcal{G}(T)$). As $X\times Y$ is complete, we conclude that $\mathcal{G}(T)$ is closed in $X\times Y$ and thus $T$ is closed.
$(\Leftarrow)$ Assume that $T$ is closed. Take a Cauchy sequence $\{u_n\}$ in $D(T)$. Then $\{(u_n, Tu_n)\}$ is Cauchy in $X\times Y$ by $(*)$. As $X\times Y$ is complete, there is $(u,v)\in X\times Y$ such that
$$\|(u_n,Tu_n)-(u,v)\|_{X\times Y}\to 0.$$
As $T$ is closed, it follows that $(u,v)\in \mathcal{G}(T)$ which implies that $u\in D(T)$ and $v=Tu$. So, $\{u_n\}$ converges to $u$ in $D(T)$ and thus $D(T)$ is complete. $\blacksquare$
