## 3 The Method
### Preliminaries and Results for the Minimising Problem Associated with (7)
\[\left\|f_{n}\right\|_{1,p(x)}\leq C. \tag{14}\]  
Indeed, since \(f_{n}\) is a minimising sequence of \(\Phi\), then we specifically have  
\[\Phi(f)\leq\alpha+\frac{1}{n},\qquad\text{ for all }n\in\mathbb{N}\setminus\{0\}. \tag{15}\]  
Consequently, by the definition of \(\Phi\) given in (12), we successively deduce  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]0and  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]1  
By using the triangular inequality and (17), it follows that there exists a constant \(K_{1}>0\), such that  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]2  
On the other hand, since \(p^{+}:=\sup_{x\in\Omega}p(x)<2\), the embedding of \(L^{2}(\Omega)\) into \(L^{p(x)}(\Omega)\) is continuous (i.e., there exists a constant \(\tilde{C}>0\) such that \(\|f_{n}\|_{L^{p(x)}(\Omega)}\leq\tilde{C}\|f_{n}\|_{L^{2}(\Omega)}\)). Thus, combined with (18), we find a constant \(K_{2}>0\), such that  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]3  
To continue, we observe that after combining the equivalence norm properties [11] (Theorem 2.3) with the estimate in (16), we can deduce that there exists a positive constant \(K_{3}\) such that  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]4  
Hence, let us return to the definition in (10) of the norm \(\|\cdot\|_{W^{1,p(x)}(\Omega)}\). By using the estimates from (19) and (20), we obtain  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]5  
In other words, we obtain the desired estimate (14). Moreover, since \(W^{1,p(x)}(\Omega)\) is a reflexive space, we can extract from the sequence \(f_{n}\) a sub-sequence that converges weakly in \(W^{1,p(x)}(\Omega)\). We denote this weak limit by \(f^{*}\). Since the application \(\Phi\) is strictly convex (as a sum of two convex functions on \(W^{1,p(x)}(\Omega)\)) and continuous for the topology of the norm \(\|\cdot\|_{W^{1,p(x)}(\Omega)}\), we can apply [12] (Corollary 3.9). In this case, we obtain  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]6  
Consequently, from the existence of \(\inf_{f\in W^{1,p(x)}(\Omega)}\), \(\Phi(f)=\alpha\) leads to \(f\) on one side  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]7  
On the other side, taking into account that the sequence \(f_{n}\) has the property from (15), this implies that, after passing to the lower limit on \(n\), the following yields  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]8  
Therefore, it is possible to conclude that  
\[L^{p(\cdot)}(\Omega):=\{u:\Omega\to\mathbb{R};\text{ measurable in }\Omega\text{ and }\int_{\Omega}|u(x)|^{p(x)}dx<+\infty\}\]9  
Then, estimates from (21) and (22) lead to the conclusion that  
\[\left\|u\right\|_{p(x)}=\inf\Biggl{\{}s>0;\,\int_{\Omega}\left|\frac{u(x)}{s} \right|^{p(x)}dx\leq 1\Biggr{\}}. \tag{9}\]0  
which establishes that the minimisation problem has a solution in \(W^{1,p(x)}(\Omega)\). The uniqueness of \(f^{*}\) is due to the strict convexity property of the function \(\Phi\). Indeed, let us reason by the opposite. This means that, suppose that the application \(\Phi\) reachesits minimum in \(f_{1}\)* and that \(f_{2}\)* has two distinct elements.