Let $(X,F,\mu)$ a measure space with $\mu(X)<\infty$, it is easy to show that, for every $u$ measurable, $$\limsup_{p \to \infty} ||u||_p \le ||u||_{\infty}$$
(it follows immediatly from the well-known inequality $||u||_p\le||u||_{\infty}\mu(X)^{1/p}$.) When $X$ has not finite measure it is easy to find a counterexample to the previuos inequality, for istance, when $X = \mathbb{R}$ dowed with the Lebesgue measure, we have $u_0\equiv 1$ and $$ 1=||u_0||_{\infty} < ||u_0||_p= + \infty. $$ My question is: is it possibile to find a (infinite measure) measure space $X$ and a $u_0$ measurable such that $$ ||u_0||_{\infty} < \limsup_{p\to\infty}|u_0||_p <+ \infty ? $$
Answer to the original question: Yes, how about: $X=\Bbb R$ with Lebesgue measure, and $u(x) = |x|^{-2/p}1_{\{|x|>1\}}$. So $\|u\|_\infty = 1$ and $\|u\|_p =2$.
Answer to the edited question: No, that is not possible. Assume $\limsup_{p\to\infty} \|u\|_p<\infty$ so that $\|u\|_p<\infty$ for some $p\ge 1$ which is fixed from now on. If $r>p$ then you always have $\|f\|_{r} \leq \|f\|_\infty^{1-\frac pr} \|f\|_p^{\frac pr}$. That follows from log convexity of Lp norms, see Lemma 9 in these notes. Letting $r\to\infty$ you get that $\limsup_{r\to\infty} \|f\|_r \leq \limsup_{r\to\infty}\|f\|_\infty^{1-\frac pr} \|f\|_p^{\frac pr} = \|f\|_\infty.$