$L_2$ integrability of a function

Question

Let $g(x,u)\in C^{\infty}([a,b] \times \mathbb{R})$

Suppose $\bar u \in L^{\infty}(a,b)$. Then, the book that I am reading says that $g(x,\bar u(x)) \in L^2(a,b)$

But, why is that true?

Answer 1

Let $m=\inf \bar{u}, M=\sup \bar{u}$. Since $g\in L^\infty\left( [a, b]\times [m, M]\right)$, the composed function $g(\cdot, \bar{u}(\cdot))$ is in $L^\infty(a, b)$.