I have a problem that I can't solve because I've just started using mathematica and matlab programs: $\mathbb{R}^2$ is Hibert space endowed with the norm$$\left\Vert x\right\Vert = \left\Vert x_{1},x_{2}\right\Vert =\left(\sum\limits_{k=1}^{2}\mid{x_{k}\mid}\right)^{\frac{1}{2}}.$$$C=[0,1]\times[0,1]$ is a closed convex subset of $\mathbb{R}^2$. Define $T$ $:C\rightarrow C$ by $T(x_{1},x_{2})=((0.01)\cos x_{1},0.01+0.02\sin x_{2})$. My purpose is to show that T is $(\alpha,\beta)$-relaxed cocoercive mapping that is $T$ satisfies:
\begin{equation*} \left\langle Tx-Ty,x-y\right\rangle \geq -\alpha ||Tx-Ty||^{2}+\beta ||x-y||^{2},\forall x,y\in C. \end{equation*} for constants $\alpha>0,\ \beta >0.$ How can I show that our operator satisfies the above inequality by using the Mathematica software?
If you have Mathematica and want to prove an inequality such as:
$$ \left\langle Tx-Ty,x-y\right\rangle \geq -\alpha ||Tx-Ty||^{2}+\beta ||x-y||^{2} $$
Then I would suggest rewriting as:
$$ g(x,y,\alpha,\beta)=\left\langle Tx-Ty,x-y\right\rangle +\alpha ||Tx-Ty||^{2}-\beta ||x-y||^{2}\geq0 $$
Because you have $(x,y)\in[0,1]\times [0,1]$, consider replacing them by $\sin^2 a$ and $\sin^2 b$. or some other function that wraps itself within the allowed range.
Then, try to minimize the modified version of $g$ for some guesses $\alpha$ and $\beta$. If you prove that the minimum is larger than zero, your inequality holds.