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Let

  • $\mathbb K\in\{\mathbb C,\mathbb R\}$
  • $H$ be a $\mathbb K$-Hilbert space
  • $\mathcal D(\mathfrak a)$ be a subspace of $H$ and $\mathfrak a:\mathcal D(\mathfrak a)\times\mathcal D(\mathfrak a)\to\mathbb K$ be sesquilinear

In Definition 1.4 of Analysis of Heat Equations on Domains the authors states that $\mathfrak a$ is called

  • accretive $:\Leftrightarrow$ $$\Re\mathfrak a(u,u)\ge0\;\;\;\text{for all }u\in\mathcal D(\mathfrak a)\tag1$$
  • continuous $:\Leftrightarrow$ $\exists M\ge0$ with $$|\mathfrak a(u,v)|\le M\|u\|_{\mathfrak a} \|v\|_{\mathfrak a}\;\;\;\text{for all }u,v\in\mathcal D(A)\tag2$$ where $$\|u\|_{\mathfrak a}:=\sqrt{\Re\mathfrak a(u,u)+\|u\|_H^2} \text{ for }u\in\mathcal D(\mathfrak a)$$

Isn't this definition broken? Unless $$\Re\mathfrak a(u,u)+\|u\|_H^2\ge0 \tag3 \text{ for }u\in\mathcal D(\mathfrak a)\;,$$ $\|\cdot\|_{\mathfrak a}$ might take values in $\mathbb C$ in which case the real numbers on the left-hand side of $(2)$ are compared with complex numbers on the right-hand side.

Clearly, if $\mathfrak a$ is accretive, then the definition of continuity is well-posed.

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    Yes, if we take $\mathfrak{a}(u,v) = -2\langle u,v\rangle$, that definition has a problem. – 2017-01-10
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    @DanielFischer Is there a consistent definition for sesquilinear forms which are not accretive? – 2017-01-10
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    Well, there's the standard definition of continuity. But if $\mathfrak{a}$ isn't defined on the whole space, one probably isn't so interested in continuity with respect to the subspace topology induced by $H\times H$. Perhaps using $\lvert \Re \mathfrak{a}(u,u)\rvert$ gives you something useful when $\mathfrak{a}$ isn't accretive. – 2017-01-10

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