If $\limsup_{p\to\infty}\|u\|_p\le C$, $u\in L^p(\Omega)$ then $u\in L^{\infty}(\Omega)$?

Question

Let $|\Omega |<\infty$, $u\in L^p(\Omega)$ for all $1\le p<\infty$ and $\limsup_{p\to\infty}\|u\|_p\le C$ for a constant $C\in\mathbb{R}$. How to prove $u\in L^{\infty}(\Omega)$? It is to show, that $\|u\|_{\infty}<\infty$, but I dont know how to do that.

Answer 1

Suppose that $\mu( \{ |f| > 2C \} ) =a > 0$, then

$$\| f\|_p \geq \left( \int_{\{ |f| > 2C \}} |f(x)|^p dx\right)^{\frac{1}{p}} \geq a^\frac{1}{p} 2C $$

But $\limsup \|f\|_p = C$ and $\lim a^\frac{1}{p} 2C = 2C > C$ : contradiction.

Answer 2

Jensen's Inequality implies that $$ \left(\frac1{|\Omega|}\int_\Omega\left|f(x)\right|^p\mathrm{d}x\right)^{1/p}=\frac{\|f\|_{L^p(\Omega)}}{|\Omega|^{1/p}}\tag{1} $$ is an increasing function of $p$.

Since $$ \int_\Omega\left|f(x)\right|^p\mathrm{d}x\le|\Omega|\,\|f\|_{L^\infty(\Omega)}^p\tag{2} $$ we have that $$ \lim_{p\to\infty}\|f\|_{L^p(\Omega)} =\lim_{p\to\infty}\left(\frac1{|\Omega|}\int_\Omega\left|f(x)\right|^p\mathrm{d}x\right)^{1/p} \le\|f\|_{L^\infty(\Omega)}\tag{3} $$ Suppose that $$ \mu=\left|\left\{x:\left|f(x)\right|\ge\alpha\right\}\right|\gt0\tag{4} $$ Then $$ \begin{align} \lim_{p\to\infty}\left(\frac1{|\Omega|}\int_\Omega\left|f(x)\right|^p\mathrm{d}x\right)^{1/p} &\ge\lim_{p\to\infty}\left(\frac{\alpha^p\mu}{|\Omega|}\right)^{1/p}\\ &=\alpha\lim_{p\to\infty}\left(\frac{\mu}{|\Omega|}\right)^{1/p}\\[3pt] &=\alpha\tag{5} \end{align} $$ If $\alpha\lt\|f\|_{L^\infty(\Omega)}$, then $(4)$ is true. Therefore, $(3)$ and $(5)$ imply that $$ \lim_{p\to\infty}\|f\|_{L^p(\Omega)} =\lim_{p\to\infty}\left(\frac1{|\Omega|}\int_\Omega\left|f(x)\right|^p\mathrm{d}x\right)^{1/p} =\|f\|_{L^\infty(\Omega)}\tag{6} $$