## 3 Main Results  
In this section, we present the \((p,q)\)-Hermite-Hadamard inequality for double integrals and the refinement of Hermite-Hadamard inequalities on the interval \([a,b]\). **Theorem 6**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex continuous function on \([a,b]\) and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}f\bigg{(}\frac{qa+pb}{p+q}\bigg{)}&\leq\frac{1}{(pb-pa)^{2}} \int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}f(tx+(1-t)y)\ _{a}d_{p,q}x_{a}d_{p,q}y\\ &\leq\frac{1}{p(b-a)}\int_{a}^{pb+(1-p)a}f(x)\ _{a}d_{p,q}x_{a}d_{p,q}y\\ &\leq\frac{qf(a)+pf(b)}{p+q}.\end{array}  \tag{12}\]  
Proof. : Since \(f\) is convex on \([a,b]\), it follows that:  
\[f(tx+(1-t)y)\leq tf(x)+(1-t)f(y) \tag{13}\]  
for all \(x,y\in[a,b]\) and \(t\in[0,1]\). Taking the double \((p,q)\)-integration on both sides for (13) on \([a,pb+(1-p)a]\times[a,pb+(1-p)a]\), we obtain:  
\[ \begin{array}{ll}\int_{a}^{pb+(1-p)a}&\int_{a}^{pb+(1-p)a}f(tx+(1-t)y)\ _{a}d_{p,q}x_{a}d_{p,q}y\\ &\leq\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}\left[tf(x)+(1-t)f(y)\right]\ _{a}d_{p,q}x_{a}d_{p,q}y\\ &=(pb-pa)\int_{a}^{pb+(1-p)a}f(x)_{a}d_{p,q}x,\end{array}  \tag{14}\]  
which show the second part of (12) by using the right-hand side of the \((p,q)\)-Hermite-Hadamard's inequality. On the other hand, by Lemma 1, we have:  
\[ \begin{array}{ll}f\bigg{(}\frac{1}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^ {pb+(1-p)a}\left(tx+(1-t)y\right)\ _{a}d_{p,q}x_{a}d_{p,q}y\bigg{)}\\ &\leq\frac{1}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}f(tx+(1-t)y) \ _{a}d_{p,q}x_{a}d_{p,q}y,\end{array} \]  
and since:  
\[\frac{1}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}\left(tx+(1-t)y \right)\ _{a}d_{p,q}x_{a}d_{p,q}y=\frac{qa+pb}{p+q}.\]  
This completes the proof. **Remark 1**. : _If \(p=1\) and \(q\to 1\), then (12) reduces to (2), that is,_  
\[f\bigg{(}\frac{a+b}{2}\bigg{)}\leq\frac{1}{(b-a)^{2}}\int_{a}^{b}\int_{a}^{b}f (tx+(1-t)y)dxdy\leq\frac{f(a)+f(b)}{2}.\]  
**Corollary 1**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex continuous function on \([a,b]\) and \(0<q<p\leq 1\). Then, we have:_  
\[f\left(\frac{qa+pb}{p+q}\right)\leq\frac{1}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a} \int_{a}^{pb+(1-p)a}f\left(\frac{x+y}{2}\right)\ _{a}d_{p,q}x_{a}d_{p,q}y\]  
\[\leq\frac{1}{p(b-a)}\int_{a}^{pb+(1-p)a}f(x)\ _{a}d_{p,q}x\]  
\[\leq\frac{qf(a)+pf(b)}{p+q}. \tag{15}\]  
**Remark 2**. : _If \(p=1\) and \(q\to 1\), then (15) reduces to:_  
\[f\left(\frac{a+b}{2}\right)\leq\frac{1}{(b-a)^{2}}\int_{a}^{b}\int_{a}^{b}f \left(\frac{x+y}{2}\right)dxdy\leq\frac{1}{b-a}\int_{a}^{b}f(x)dx\leq\frac{f(a) +f(b)}{2},\]  
_which readily appeared in [25]._  
**Theorem 7**. : _Let \(f:[a,b]\rightarrow\mathbb{R}\) be a convex continuous function on \([a,b]\) and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{split}\frac{p}{(pb-pa)^{2}}&\int_{a}^{pb+(1 -p)a}\int_{a}^{pb+(1-p)a}f\left(\frac{px+qy}{p+q}\right)\,_{a}d_{p,q}x_{a}d_{p,q }y\\ &\leq\frac{1}{(pb-pa)^{2}}\int_{0}^{p}\int_{a}^{pb+(1-p)a}\int_{a }^{pb+(1-p)a}f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y_{a}d_{p,q}t\\ &\leq\frac{1}{pb-pa}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x.\end{split}  \tag{16}\]  
Proof. : Let \(g:[a,b]\rightarrow\mathbb{R}\) be given by:  
\[g(t)=\frac{1}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}f(tx+(1-t)y )\,_{a}d_{p,q}x_{a}d_{p,q}y.\]  
For all \(t_{1},t_{2}\in[0,1]\) and \(\alpha,\beta\geq 0\) with \(\alpha+\beta=1\), we consider:  
\[ \begin{split} g(\alpha t_{1}+\beta t_{2})&=\frac{1 }{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}f((\alpha t_{1}+\beta t_ {2})x+(1-(\alpha t_{1}+\beta t_{2}))y)\,_{a}d_{p,q}x_{a}d_{p,q}y\\ &\leq\frac{\alpha}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+ (1-p)a}f(t_{1}x+(1-t_{1})y)\,_{a}d_{p,q}x_{a}d_{p,q}y\\ &\quad+\frac{\beta}{(pb-pa)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb +(1-p)a}f(t_{2}x+(1-t_{2})y)\,_{a}d_{p,q}x_{a}d_{p,q}y\\ &=\alpha g(t_{1})+\beta g(t_{2}),\end{split} \]  
which show that \(g\) is convex on \([0,1]\). Using Theorem 3 for the convex function \(g\), we have:  
\[