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: 
In general the implication $u\in L^p(\Omega) \implies u \in L^\infty(\Omega)$ is false. 

For example, if the space $\Omega$ has finite measure (e.g. $\Omega=[0,1]$) it is the converse to hold: 

$$u\in L^\infty(\Omega) \implies u \in L^p(\Omega) \quad \forall p\in [1,\infty],$$

and more in general:

$$u\in L^q(\Omega) \implies u \in L^p(\Omega) \quad \forall 1\le p \le q \le \infty.$$

One counterexample is simply $f(x)=\frac 1{\sqrt x},$ for $x\in (0,1):$ it belongs to $L^1((0,1))$ but not to $L^\infty((0,1)).$

If you look for something similar to the converse to this last statement for general $\Omega$ there is this implication: 
$$f\in L^\infty(\Omega) \implies \left(f\in L^p \implies f\in L^q\right) \quad \forall 1\le p\le q\le \infty.$$
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.)