Let be $\Omega \subset \mathbb{R}^n$ an open set and $E = H_0^1(\Omega) \cap L^p(\Omega)$ with $1 \leq p < \infty$ provided with the norm $||u||_E = ||u||_{H_0^1(\Omega)} + ||u||_{L^p(\Omega)}$. Show that $(E,||\cdot||_E)$ is a Banach space.
I would like to know if my attempt is right.
$\textbf{My attempt:}$
Let be $(\varphi_m) \subset E$ a Cauchy sequence, then there are $u \in H_0^1(\Omega)$ and $v \in L^p(\Omega)$ such that $||u - \varphi_m||_{H_0^1(\Omega)} \rightarrow 0$ and $||v - \varphi_m||_{L^p(\Omega)} \rightarrow 0$. These functions are in $L_{\text{loc}}^1(\Omega)$ because $u \in L^2(\Omega)$ and $v \in L^p(\Omega)$. We only need to show that $||u - v||_{L_{\text{loc}}^1(\Omega)} = 0$ to ensure $u \in E$, but this occurs because
$\begin{align*} 0 &\leq ||u - v||_{L_{\text{loc}}^1(\Omega)}\\ &\leq ||u - \varphi_m||_{L_{\text{loc}}^1(\Omega)} + ||\varphi_m - v||_{L_{\text{loc}}^1(\Omega)}\\ &\overset{m \rightarrow \infty}{\rightarrow} 0, \end{align*}$
where this convergence is true from the convergences previous and because $L^p(\Omega) \hookrightarrow L_{\text{loc}}^1(\Omega)$ for $1 \leq p < \infty$. $\square$
Thanks in advance!
Your attempt is flawed because $L^1_{loc}$ is not a normed space. Instead take arbitrary compact $K\subset \Omega$. Then the convergence of $\phi_m$ implies $\|u-\phi_m\|_{L^1(K)}\to 0$ and $\|v-\phi_m\|_{L^1(K)}\to 0$.
Here is a different attempt without talking about $L^1_{loc}$.
We have $\phi_m \to u$ in $H^1_0$ and $\phi_m \to v$ in $L^p$. It only remains to show $u=v$. Take $z\in C_c^\infty(\Omega)$. Then $\phi\mapsto \int_\Omega z\phi $ is a linear and bounded functional on both $H^1_0$ and $L^p$. This proves $$ \int_\Omega \phi_m z \to \int_\Omega u z $$ and $$ \int_\Omega \phi_m z \to \int_\Omega v z. $$ It follows $\int_\Omega (u-v)z=$ for all $z\in C_c^\infty(\Omega)$. Hence $u=v$ a.e.