# Proving the graph norm is indeed a norm

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