Question: I want to verify when $u(x) \in L^p$, $x \in [0, \infty)$

then can we say $u(x) \in L^{ \infty}$, $x \in [0, \infty)$ ?

Where $u$ is any function with variable x, and $L^p$ is talking about Lebesgue spaces.

If not, please give me an counterexample 

or we could derive that result by adding any conditions?

Answer: 
No. For any $1 \le p < +\infty$ set $u(x) = \frac1{x^{1/2p}}\chi\_{(0,1)}(x)$. Then

$$\int\_0^\infty|u(x)|^p\,dx = \int\_0^1 \frac{dx}{\sqrt{x}} < +\infty$$

so $u \in L^p(0, \infty)$. However, for any $M \ge 1$ we have $u \ge M$ on $\left(0, \frac1{M^{2p}}\right)$ which is a set of positive measure so $u \notin L^\infty(0,\infty)$.
No. In fact for a given measure space the inclusion $L^p\subset L^\infty$ holds if and only if $$\inf\{\mu(E):\mu(E)>0\}>0.$$(Which holds for counting measure, for example.)