Proving the graph norm is indeed a norm

Question

I was reading p.238-239 in the book enter link description here

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?

Answer 1

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:

1. $\|u\|_T> 0$ if $u\neq 0$;
2. $\|a u\|_T=|a|\|u\|_T$ for any scalar $a$;
3. $\|u+v\|_T\leq\|u\|_T+\|v\|_T$ 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:

$$\begin{align} \|u+v\|_T^2&=\|u+v\|^2_X+\|T(u+v)\|^2_X \qquad\text{(by definition of $\|\cdot\|_T$)}\\\\ &\leq (\|u\|_X+\|v\|_X)^2+(\|Tu\|_X+\|Tv\|_X)^2\qquad\text{(by triangle inequality)}\\\\ &= \|u\|_X^2+2\|u\|_X\|v\|_X+\|v\|_X^2+\|Tu\|_X^2+2\|Tu\|_X\|Tv\|_X+\|Tv\|_X^2\\\\ &\leq\|u\|^2_X+\|Tu\|^2_X+2\sqrt{(\|u\|^2_X+\|Tu\|^2_X)(\|v\|^2_X+\|Tv\|^2_X)}+\|v\|^2_X+\|Tv\|^2_X\\\\ &=\|u\|_T^2+2\|u\|_T\|v\|_T+\|v\|_T^2\\\\ &=(\|u\|_T+\|v\|_T)^2 \end{align}$$

Answer 2

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 :

$$ |v|_T|w|_T =\sqrt{|v|^2|w|^2 + |Tv|^2|Tw|^2+ |v|^2|Tw|^2 + |w|^2|Tv|^2 } $$

$$ 2|v||w||Tv||Tw| \leq |v|^2|Tw|^2 + |w|^2|Tv|^2 $$