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I'm trying to solve below exercise in Brezis's Functional Analysis, i.e.,

Let $(\Omega, \mathcal F, \mu)$ be a finite measure space. Assume $f \in \bigcap_{p \in [1, \infty)} L^p (\Omega)$ and there is $C \in [0, \infty)$ such that $\|f\|_p \le C$ then $f \in L^\infty(\Omega)$.

  • Could you confirm if my below attempt is correct?

  • Could you confirm if the assumption $\mu(\Omega) < \infty$ is not needed?

Proof Fix $K> C$. We have $$ \mu (\{|f| \ge K\}) =\int 1_{\{|f|^p \ge K^p\}} \mathrm d \mu \le \int \frac{|f|^p}{K^p} 1_{\{|f|^p \ge K^p\}} \mathrm d \mu \le \bigg ( \frac{\|f\|_p}{K} \bigg)^p \le \bigg ( \frac{C}{K} \bigg)^p. $$

Taking the limit $p \to \infty$, we get $$ \mu (\{|f| \ge K\}) =0. $$

Hence $\|f\|_\infty \le K$ and thus $\|f\|_\infty \le C$. This completes the proof.

Akira
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    So you assume that the bound $\lVert f\rVert_p\leq C$ is uniform in $p\geq1$? In that case I see no problems with your solution. – jakobdt Mar 30 '23 at 11:00
  • @jakobdt oh yess, the bound is uniform in $p$. Thank you so much for your verification! – Akira Mar 30 '23 at 11:03
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    Finiteness is not require. It all follows from the fact that $\lim_{p\rightarrow\infty}|f|p=|f|\infty$. The assumption in your posting means that $|f|_\infty\leq C$. – Mittens Mar 30 '23 at 16:53

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