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How are $C^0,C^1$ norms defined? I know $L_p,L_\infty$ norms but are the former defined.

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  • $\begingroup$ In $C^0[a,b]$ one usually uses the $\sup$ norm, $$\lVert f-g\rVert_\infty=\sup_{x\in [a,b]}|f(x)-g(x)|$$ Are you meaning to ask about $C^0[a,b]$? $\endgroup$
    – Pedro
    Commented Aug 8, 2013 at 16:05
  • $\begingroup$ @PeterTamaroff, I think so. How about $C^1$ $\endgroup$
    – Vaolter
    Commented Aug 8, 2013 at 16:08

2 Answers 2

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On $\mathcal{C}^{0}([a,b])$, the usual norm is

$$ \Vert f \Vert = \sup \limits_{x \in [a,b]} \vert f(x) \vert $$

(the interesting point is that $\left( \mathcal{C}^{0}([a,b]), \Vert \cdot \Vert \right)$ is a Banach space.

On $\mathcal{C}^{1}([a,b])$, you can define the norm

$$ \Vert f \Vert_{\mathcal{C}^{1}} = \sup_{x \in [a,b]} \vert f(x) \vert + \sup_{x \in [a,b]} \vert f'(x) \vert $$

(and $\left( \mathcal{C}^{1}([a,b]),\Vert \cdot \Vert_{\mathcal{C}^{1}} \right)$ is also a Banach space.)

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Let $\Omega$ be a domain.

Set $$[u]_{k,0,\Omega} = \sup_{|\beta| =k}\sup_{\Omega}|D^{\beta}u|$$

Then, $$||u||_{C^{k}(\bar{\Omega})} = \sum_{j=0}^k{[u]_{j,0,\Omega}}$$

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