Let $f\in L^{\infty}(\mathbb{R}^n)$. It is well known from functional analysis that $ f(x) \leq ||f(x)||_{\infty} $ for almost everywhere $x\in \mathbb{R}^n.$ I suspect that if $f$ is continuous, this inequality holds for all points $x\in \mathbb{R}^n$, precisely, since locally f is bounded. Is this true? How can we justify?
Asked
Active
Viewed 34 times
0
-
For a point that violates the inequality, use continuity to find ... – user10354138 Sep 21 '23 at 19:19
-
@user10354138 finds a neighborhood (the open ball) where f applied at such points remains violating the inequality, this means that there will be a set of positive measures that violate the inequality, this contradicts that $|f(x)| \leq ||f||_{\infty}$ almost everywhere? – Ilovemath Sep 21 '23 at 19:51
-
1https://math.stackexchange.com/questions/618101/essential-supremum-with-the-continuous-function – Evangelopoulos Foivos Sep 23 '23 at 19:26