# Proving the graph norm is indeed a norm

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?

• Take a screenshot of your link; it's inaccessible for some of us. – parsiad Jan 6 '16 at 5:26
• I think that Kato's "Perturbation theory of linear operators" contains all required material. – Urgje Jan 6 '16 at 9:32
• @Urgie Do you know what page (or what chapter) in Kato? – math101 Jan 6 '16 at 11:43

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}

• Thanks, both of the attempts. My apology that the question I have in mind is in the context of Banach space. Hope you guys could help... – math101 Jan 6 '16 at 6:01
• @math101 See remarks 2 and 3. – Pedro Jan 6 '16 at 6:16
• what if I consider dense subspace of $L^p$ rather than $L^2$, the inner product structure is not relevant to the context I have in mind – math101 Jan 6 '16 at 6:53
• @math101 Sorry, I didn't understand your doubt. – Pedro Jan 6 '16 at 7:03
• Perhaps I shouldn't mention the book, what I asked for is just what I've written. I have edited my question to clarify. – math101 Jan 6 '16 at 7:07

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 :

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

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

• I was asking a question in the context of Banach space. Sorry that I forgot to emphasis this... – math101 Jan 6 '16 at 6:13
• I do not know well about the context. But I think that the first line in 239p. means that if $(V_i, |\ |_i)$ is normed space then $(V_1\times \cdots \times V_n, |\ |)$ is a normed space where $|(v_1,\cdots, v_n)|^2 =\sum_{i=1}^n |v_i|_i^2$ – HK Lee Jan 6 '16 at 6:22
• @math101 The HKLee's answer is general (However, it seems that there is a typo). The part "Not considering Banach" means that the completeness is not needed (I guess). – Pedro Jan 6 '16 at 6:24
• @math101 Sorry. I must consider the completeness (as Pedro said) I will edit. – HK Lee Jan 6 '16 at 6:28
• Thanks both of you. I will learn about what you will write – math101 Jan 6 '16 at 6:28