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I read a paper, where the derivative operator is made continuous by setting up a few assumptions. I thought, it would be enough to extend the sup-norm to $\|\cdot\|_{\infty,\infty} :=\|\cdot\|_\infty + \|\nabla_x \cdot\|_\infty$ to achieve that (like here on SE). But the authors do additional things too.

Let $\Omega \subset \mathbb{R}^d$ be a compact, convex set. We say the spatial derivative $\nabla_x\phi$ has a continuous extension to $\Omega$, if there exists $g\in C(\Omega,\mathbb{R}^{k,d})$ such that $g|_{\Omega^\circ} = \nabla_x\phi$.

Let further

$$C^1(\Omega,\mathbb{R}^k) = \{\phi:\mathbb{R}^d \rightarrow \mathbb{R}^k | \phi \text{ continuously differentiable on } \Omega^\circ , \\ \nabla_x\phi \text{ has a continuous extension to }\Omega, \\||\phi||_{\infty,\infty} = \|\phi\|_\infty + \|\nabla_x\phi\|_\infty < \infty \}$$

It is said, that this makes $\nabla_x$ a continuous linear operator. Why is this continuous extension necessary?

dba
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  • Derivatives are defined on open sets, and $\Omega$ is compact. – cmk Aug 19 '21 at 13:14
  • I dont see how this matters here. But for the example in the linked thread, which was $C^1([0,1])$, this did not seem to matter.since the chosen norm guarantees a bounded gradient. If I extend the derivative to the boundary or not, does not change the boundedness of $\nabla_x$, or does it? – dba Aug 19 '21 at 13:36

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