I am reposting a question from Math Overflow, because it seems it gets no attention.
Let $\Omega\subset \mathbb R^{d=3}$ is a bounded and Lipschitz domain. Let $u\in H_g^1(\Omega)$ satisfy the weak formulation $$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)\quad\quad (*)$$ where $a(u,v)=\int\limits_{\Omega}{\nabla u\cdot\nabla vdx}$, $f(u)=\sinh (u)=\frac{e^u-e^{-u}}{2}$, and $H_g^1(\Omega)=\{u\in H^1(\Omega):\,u=g \text{ on } \partial \Omega\}$.
Note that there is no polynomial growth condition on the nonlinearity $f$ and so $f(u)$ is not supposed to be in $L^2(\Omega),\,\forall u\in H_g^1(\Omega)$. In $d=3$, there are functions $u\in H^1(\Omega)$ s.t $e^u\notin L^1(\Omega)$ (for instance $u=\ln\left(\frac{1}{r^\alpha}\right)\in H_0^1(B_1),\,\alpha\ge 3$ over the unit ball $B_1$).
Question: Is it true that equation $(*)$ is satisfied for all $v\in H_0^1(\Omega)$ ? (If there is an a priori estimate which gives $u\in L^\infty(\Omega)$ then $(∗)$ is obviously satisfied for all $v\in H_0^1(\Omega)$)
If it would help, we may assume that $f(u)\in L^1(\Omega)$.
If I have a positive answer to the question, this means that equation $(*)$ has a unique solution: suppose $u_1,u_2$ are solutions to $(*)$, then $a(u_1-u_2,v)+(f(u_1)-f(u_2),v)=0,\,\forall v\in H_0^1(\Omega)$ and so we can test with $u_1-u_2\in H_0^1(\Omega)$ and use the monotonicity property $(f(u_1)-f(u_2),u_1-u_2)\ge 0$ to conclude that $u_1=u_2$.
My try: From equation $(*)$ we see that $\left|\int\limits_{\Omega}{f(u)vdx}\right|\leq \|u\|_1\|v\|_1,\,\forall v\in H_0^1(\Omega)\cap L^\infty(\Omega)$ and therefore $f(u)$ defines a bounded linear functional over the dense subspace $H_0^1(\Omega)\cap L^\infty(\Omega)$ by the formula $\langle F,v\rangle=\int\limits_{\Omega}{f(u)vdx}$ and so it is uniquely extendable over the whole $H_0^1(\Omega)$ by continuity to the functional $\bar F\in H^{-1}(\Omega)$. However, the extension is not necessarily reprezentable by the same formula. Now, let $v\in H_0^1(\Omega)$ is arbitrary and $\{v_n\}\subset H_0^1(\Omega)\cap L^\infty(\Omega)$ s.t $$\begin{array}{|l} &v_n(x)\to v(x),\,a.e\\ & \|v_n-v\|_1\to 0 \end{array}$$ Then, $\int\limits_{\Omega}{f(u)v_ndx}=-a(u,v_n),\,\forall n$ and $a(u,v_n)\to a(u,v)\Rightarrow \int\limits_{\Omega}{f(u)v_ndx}\to -a(u,v)=:A$. Now, I want to conclude that $A=\int\limits_{\Omega}{f(u)vdx}$. We also have $f(u(x))v_n(x)\to f(u(x))v(x),\,a.e$ in $\Omega$ and $\int\limits_{\Omega}{f(u)v_ndx}=\langle F,v_n\rangle\to \langle \bar F,v\rangle$. But from here I still cannot show that $\int\limits_{\Omega}{f(u)v_ndx}\to \int\limits_{\Omega}{f(u)vdx}$.
I cannot use Lebesgue DCT, because I do not have a dominating function for $|f(u(x))v_n(x)|$. I thought also about the generalized Lebesgue DCT and maybe Vitali convergence theorem, but couldn't do it.
If we can use some type of regularity theory or maximum principle (which I do not know for this equation) and show that $u\in L^\infty(\Omega)$ or just that $f(u)\in L^{6/5}(\Omega)$, then $$\left|\int\limits_{\Omega}{f(u)(v_n-v)dx}\right|\leq \|f(u)\|_{L^{6/5}(\Omega)}\|v_n-v\|_{L^6(\Omega)}\leq C\|f(u)\|_{L^{6/5}(\Omega)}\|v_n-v\|_1\to 0$$ and I would be done.
You can show your regulary if $g \in W^{1,p}(\Omega)$, $p > d \ge 2$, by using a trick by Stampacchia.
First, we note $$a(u - g,v) + \int_\Omega f(u) v \, \mathrm{d} x = a(-g,v).$$
Now, let $K > 0$ and set $$(u - g)_K = \begin{cases} u - g & \text{if } |u - g| \le K, \\ \operatorname{sign}(u-g) \, K & \text{else}.\end{cases}$$ Then, $(u-g)_K \in H_0^1(\Omega) \cap L^\infty(\Omega)$.
Now, we choose $v = (u-g)_K$ and use $$a(u-g, (u-g)_K) = a((u-g)_K,(u-g)_K)$$ and $$\int_\Omega f(u) \, (u-g)_K \, \mathrm{d}x = \int_\Omega (f(u) - f(g)) \, (u-g)_K \, \mathrm{d}x + \int_\Omega f(g) \, (u-g)_K \, \mathrm{d}x \\\ge \int_\Omega f(g) \, (u-g)_K \, \mathrm{d}x$$ by monotonicity of $f$.
This shows $$ | (u-g)_K |_{H_0^1}^2 \le -\int_\Omega \nabla g \nabla (u-g)_K \, \mathrm{d}x - \int_\Omega f(g) \, (u - g)_K \, \mathrm{d}x$$ for all $K > 0$ and $\nabla g \in L^p(\Omega)$, $f(g) \in L^\infty(\Omega)$.
This estimate is enough to conclude $u - g \in L^\infty(\Omega)$. The detailed arguments can be found in many books, e.g., "An introduction to variational inequalities and their applications" by Kinderlehrer and Stampacchia, proof of Theorem B.2.
It is a remarkable feature of this trick that only the monotonicity of $f$ is used.