Let $g\in C^\infty ([a,b]\times\mathbb{R})$, $(x,u)\mapsto g(x,u)$. Does there exist a $\beta>0$ such that $\left|\frac{\partial g(x,u)}{\partial u}\right|\le \beta(1+|u|)$?
If not, how can I justify that $\int_a^b \frac{\partial g(x,u)}{\partial u} \varphi(x)dx$ converges, where $\varphi$ represents an element in $W_0^{1,2}(a,b)$ and where $a,b\in\mathbb{R}$ with $b\ge a$? Here in the intgral we consider $u$ as a representative of a function in $W^{1,2}(a,b)$ (i.e. we have $g(x,u(x))$).
If there exists such a $\beta$, then you can estimate as follows: $\int_a^b \left|\frac{\partial g(x,u)}{\partial u}\right|^2dx\le \int_a^b (\beta(1+|u|))^2dx<\infty$ for $u\in W^{1,2}(a,b)$ and therefore $\left|\frac{\partial g}{\partial u}\right|_{L^2}<\infty$. It follows with Hölder:
$$\int_a^b\left| \frac{\partial g(x,u)}{\partial u} \varphi(x)\right|dx\le \|\frac{\partial g}{\partial u}\|_{L^2}\|\varphi\|_{L^2}<\infty$$
$W_0^{1,2}(a,b)$ is defined as the closure of $C_0^\infty(a,b)$ (smooth functions with compact support in $(a,b)$ with respect to the $W^{1,2}$-norm.
Thank you.
The function $g(x,u)=e^u$ is $C^\infty$ but does not satisfy your inequality. If $u\in H^1_0(a,b)$ then $u$ has a representative which is continuous in $[a,b]$. Hence, if you consider the compact set $K=[a,b]\times [-M,M]$, where $M=\max_{[a,b]}|u|$, you have that $\frac{\partial g}{\partial u}$ is bounded in $K$ and so $\int \frac{\partial g}{\partial u}(x,u)\varphi$ makes sense.