If $\limsup_{p\to\infty}\|u\|_p\le C$, $u\in L^p(\Omega)$ then $u\in L^{\infty}(\Omega)$? 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.
 A: 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.
A: 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
$$
\lim_{p\to\infty}\|f\|_{L^p(\Omega)}
=\lim_{p\to\infty}\left(\frac1{|\Omega|}\int_\Omega\left|f(x)\right|^p\mathrm{d}x\right)^{1/p}
=\|f\|_{L^\infty(\Omega)}\tag{6}
$$
