= Solution
Choose the real <zero-trace Sobolev space>
$$
X=W_0^{1,4}(\Omega),\qquad\|u\|_X=\|\nabla u\|_4,
\qquad Y=L^{8/7}(\Omega).
$$
The <Poincaré inequality> makes this an equivalent <Sobolev norm>. Since the dimension is five, the <Sobolev conjugate exponent> is $4^*=5\cdot4/(5-4)=20$. The <Sobolev embedding theorem> gives $X\hookrightarrow L^{20}(\Omega)$ continuously. Boundedness of $\Omega$ and <Lp inclusion on a finite measure space> then give $X\hookrightarrow L^8(\Omega)$ continuously, say $\|u\|_8\leq C_8\|u\|_X$.
For $f\in Y$, the <weak formulation> is
$$
\boxed{u\in X,\qquad
\int_\Omega |\nabla u|^2\nabla u\cdot\nabla v\,dx
+\int_\Omega u^7v\,dx=\int_\Omega fv\,dx
\quad\forall v\in X.}
$$
The zero <Dirichlet boundary condition> is imposed through the <Sobolev trace>. <Hölder's inequality> verifies every term:
$$
\begin{aligned}
\left|\int_\Omega|\nabla u|^2\nabla u\cdot\nabla v\,dx\right|
&\leq\|\nabla u\|_4^3\|\nabla v\|_4,\\
\left|\int_\Omega u^7v\,dx\right|
&\leq\|u\|_8^7\|v\|_8,\\
\left|\int_\Omega fv\,dx\right|
&\leq\|f\|_{8/7}\|v\|_8.
\end{aligned}
$$
Thus $|\nabla u|^2\nabla u\in L^{4/3}(\Omega;\mathbb R^5)$, while $u^7\in L^{8/7}(\Omega)$, and all three pairings are finite bounded <linear functionals> in $v$. The choice $Y=L^{8/7}(\Omega)$ is sufficient; it is not intended to be the largest possible space of data.
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