Let $D$ be an open and bounded set in $R^d$ with Lipschitz boundary. Denote by $C(\bar{D})$, the space of continuous functions $f : \bar{D} → R$. When equipped with the supremum norm $$\|f\|_{C(\bar{D})} = \sup_{x\in D} |f(x)|, $$ $C(\bar{D})$ is a Banach space. What is the difference with the $L^\infty(D)$ space with norm $$\|f\|_{L^\infty(D)}=\mathrm{ess}\,\sup |f(x)| $$

  • $\begingroup$ The first is a closed subspace of the latter. $\endgroup$ – Giuseppe Negro Jan 13 '16 at 14:39
  • $\begingroup$ ??? The elements of $L^\infty$ are just measurable, not continuous. $\endgroup$ – David C. Ullrich Jan 13 '16 at 14:48
  • $\begingroup$ I see, but are the two norms equal $\|f\|_{C(\bar{D})}=\|f\|_{L^\infty(D)}$? $\endgroup$ – sjage Jan 13 '16 at 14:55

As continuous on a compact sets, all the functions in $C(\overline{D})$ belong to $L^{\infty}(D)$ (note that the boundary has measure zero), so $C(\overline{D})\subset L^{\infty}(D)$. But, as the functions in $L^{\infty}$ need not to be continuous, $C(\overline{D})$ is a proper subspace of $L^{\infty}(D)$.

The two norms coincide on $C(\overline{D})$, as in general $\mathrm{ess}\,\sup |f(x)| \leq \sup |f(x)|$ and equality holds by continuity.


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