Proving the graph norm is indeed a norm I was reading p.238-239 in the book
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Example A.14 in that book motivates me to think of the following question.
And have trouble verifying the following fact from functional analysis.
Consider operators of the form $T: V\to \mathcal{E},$ where $V\subset E$ is a (not necessarily closed) linear subspace, which are linear. We say that $T$ is densely defined if $V$ is dense in $\mathcal{E}$. We say that $T$ is closed if the graph $\{(v,Tv):v\in V\}$ is a closed subspace of $\mathcal{E}\oplus\mathcal{E}.$
Let $T$ be the operator in Banach space $\mathcal{E}$ with the domain $D(T).$
The graph norm on $D(T)$ is the norm is defined by
$$\|v\|_T=\|v\|_{\mathcal{E}}+\|Tv\|_{\mathcal{E}}$$ for all $v\in D(T).$
How to show the graph norm above is indeed a norm on $D(T)?$ Here, $D(T)$ is the set of $\phi\in L^p(X)$ for which $A\phi$ exists.
Does anyone have a solution of the question I asked above?
 A: I think that this may be followed from direct computations by using definition of norm and inequality formula, i.e., Cauchy-Schwartz. 
(1) Since $T$ is linear so $|0|_T^2 = |0|^2 + |T0|^2 =0 $ That is $|0|_T=0$ If $|v|_T=0$, then $|v|=0$ Hence $v=0$
(2) $|cv|_T^2 =c^2|v|^2 + c^2|Tv|^2 $ so that $|cv|_T^2=|c||v|_T$
(3) $$|v+w|_T^2 =|v+w|^2 + |Tv+Tw|^2 \leq (|v|+|w|)^2 + (|Tv|+|Tw|)^2
 = |v|_T^2 + |w|_T^2 +2|v||w| + 2|Tv||Tw|$$
So we have a claim $$ |v||w| + |Tv||Tw| \leq |v|_T|w|_T $$
Note that this is followed from these : 
$$ |v|_T|w|_T =\sqrt{|v|^2|w|^2 + |Tv|^2|Tw|^2+ |v|^2|Tw|^2 + |w|^2|Tv|^2 } $$
$$ 2|v||w||Tv||Tw| \leq  |v|^2|Tw|^2 + |w|^2|Tv|^2  $$
A: I will use $X$ instead of $\mathcal{E}$.
Let $X$ be a normed space and $T:D(T)\to X$ a linear operator, where $D(T)$ is a linear subspace of $X$. The graph norm on $D(T)$ is given by
$$\|u\|_T^2=\|u\|_X^2+\|Tu\|_X^2$$

How to show the graph norm above is indeed a norm on $D(T)?$

We have to verify the following conditions:


*

*$\|u\|_T> 0$ if $u\neq 0$;

*$\|a u\|_T=|a|\|u\|_T$ for any scalar $a$;

*$\|u+v\|_T\leq\|u\|_T+\|v\|_T$ for any vectors $u$ and $v$ (triangle inequality).


Proof of 1: As $\|\cdot\|_X$ is a norm, we have $\|u\|_T^2=\|u\|_X^2+\|Tu\|_X^2\geq \|u\|_X^2>0$  if $u\neq 0$ and thus 1 is valid.
Proof of 2: As $T$ is linear and $\|\cdot\|_X$ is a norm, we have 
$$\|a u\|_T^2=\|a u\|_X^2+\|T(a u)\|_X^2=|a|^2\|u\|_X^2+|a|^2\|Tu\|_X^2=|a|^2\|u\|_T^2$$
for any scalar $a$ and thus 2 is valid.
Proof of 3:
$$\begin{align}
\|u+v\|_T^2&=\|u+v\|^2_X+\|T(u+v)\|^2_X \qquad\text{(by definition of $\|\cdot\|_T$)}\\\\
&\leq (\|u\|_X+\|v\|_X)^2+(\|Tu\|_X+\|Tv\|_X)^2\qquad\text{(by triangle inequality)}\\\\
&= \|u\|_X^2+2\|u\|_X\|v\|_X+\|v\|_X^2+\|Tu\|_X^2+2\|Tu\|_X\|Tv\|_X+\|Tv\|_X^2\\\\
&\leq\|u\|^2_X+\|Tu\|^2_X+2\sqrt{(\|u\|^2_X+\|Tu\|^2_X)(\|v\|^2_X+\|Tv\|^2_X)}+\|v\|^2_X+\|Tv\|^2_X\\\\
&=\|u\|_T^2+2\|u\|_T\|v\|_T+\|v\|_T^2\\\\
&=(\|u\|_T+\|v\|_T)^2
\end{align}$$
