If $u \in C$, $|\int u \phi_{ij}| \leq 2 \|h\|_p \|\phi\|_{p'} \forall \phi \in C_c^\infty$, then $\|u_{ij}\|_p \leq 2\|h\|_p$? Let $\Omega$ be a domain in $\mathbb R^N$, $u \in C(\Omega)$ and $h \in L^p(\Omega)$. Suppose that
$$
\left|\int_\Omega u \varphi_{ij} \ dx  \right| \leq 2\|h\|_p \|\varphi\|_{p'} \quad \forall \varphi \in C_c^\infty(\Omega).
$$

Is it true that $u_{ij} \in L^p(\Omega)$ and $\|u_{ij}\|_p \leq 2\|h\|_p$? If so, how to prove it?

Context: from what I understand, Caffarelli and Cabré use this fact in the proof of Proposition 1.1 in their book Fully Nonlinear Elliptic Equations. I understand that $u_{ij}$ is to be interpreted as distributional derivatives, but I am unable to prove the inequality.
Any hints will be the most appreciated.
Thanks in advance.

Define $L : C_c^\infty(\Omega) \to \mathbf R$ by
$$L\phi = \int_\Omega u \phi_{ij} \, dx.$$ By hypothesis you have
$$|L\phi| \le M \|\phi\|_{p'},\quad \phi \in C_c^\infty(\Omega).$$ In particular $L$ is bounded (in the $L^{p'}$ norm) on $C_c^\infty(\Omega)$.  Since $C_c^\infty(\Omega)$ is dense in $L^{p'}(\Omega)$ you can extend $L$ by continuity to all of $L^{p'}(\Omega)$. If you denote the extension by $\tilde L$ then
$$ |\tilde Lg| \le M \|g\|_{p'},\quad g \in L^{p'}(\Omega).$$
Now apply Riesz representation theorem (you need to insist $1 < p \le \infty$) to find a function $f$ satisfying $\|f\|_p \le M$ and
$$\tilde L g = \int_\Omega fg \, dx, \quad g \in L^{p'}(\Omega).$$
In particular, if $\phi \in C_c^\infty(\Omega)$ then
$$ \int_\Omega u \phi_{ij} \, dx = L\phi = \tilde L\phi = \int_\Omega f \phi \, dx.$$
Consequently $f = u_{ij}$ in the distributional sense and $\|u_{ij}\|_p \le M$.
