Considering a function $f\in\mathscr({C}(K),\lvert\lvert \cdot\rvert\rvert_\infty)$, i.e.$$f:K\rightarrow\mathbb{C}$$ everywhere continuous where $K$ is a compact subset of $\mathbb{R}^n$, does $f$ lie in $L^p(K)$ for every $1\le p\le\infty$?

I would say yes. The $p=\infty$ case is true because the essential supremum is smaller or equal to the actual supremum everywhere, and for $p\not=\infty$$$\int_K\lvert f(x)\rvert^p \mathrm{d}x\le \sup_{x\in K}\lvert f(x)\rvert^p \mu(K)^p$$ where $f$ assumes its maximum because of continuity everywhere and $K$ being compact, and the Lebesgue-measure of a compact set is also finite as I understand it.

I'm not sure though, because in the solutions for an exercise I did they use $K=[0,1]^2$ and say that for $x\in K$ one has $\int_0^1\lvert f(x,y)\rvert^2\mathrm{d}y\lt\infty$ only almost everywhere, but shouldn't this hold everywhere?

I didn't write that $K$ lies in $\mathbb{R}^n$ in the title because I assume this applies for general compact measure spaces as well?

  • $\begingroup$ Hint: $|.|^p$ is continuous. $\endgroup$
    – Aloizio Macedo
    Sep 23, 2015 at 19:04
  • 1
    $\begingroup$ You are correct on all accounts. $\endgroup$ Sep 23, 2015 at 20:02
  • $\begingroup$ @JuliánAguirre That's nice to read. $\endgroup$
    – azureai
    Sep 24, 2015 at 7:15
  • $\begingroup$ @AloizioMacedo Do you mean that the whole expression can be read as a continuous function on K and hence assumes its maximum for $1\le p\le\infty$, i.e. can $\int_K(\cdot)\mathrm{d}\mu$ be interpreted as a continuos function? $\endgroup$
    – azureai
    Sep 24, 2015 at 7:16

1 Answer 1


You are right. The inequality $$\int_K\lvert f(x)\rvert^p \mathrm{d}x\le \sup_{x\in K}\lvert f(x)\rvert^p \mu(K)^p$$ shows that $C(K)\subset L^p(K)$ for every compact topological space equipped with a finite Borel measure $\mu$.


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