Let $f(x,y)=\left(\frac{1}{2}x^{T}Qx\right)\left(\frac{1}{2}y^{T}Ry\right)$.given $g(x)=f(x,x)$ not convex.does $f$ convex?

Question

I really need with this one:

Let $$\begin{array}{c} Q,R\in\mathbb{R}^{n},Q,R\succ0\\ f:\mathbb{R}^{n}\times\mathbb{R}^{n}\longrightarrow\mathbb{R},g:\mathbb{R}^{n}\longrightarrow\mathbb{R}\\ f\left(\boldsymbol{x},\boldsymbol{y}\right)=\left(\frac{1}{2}\boldsymbol{x}^{T}Q\boldsymbol{x}\right)\left(\frac{1}{2}\boldsymbol{y}^{T}R\boldsymbol{y}\right)\\ g\left(\boldsymbol{x}\right)=f\left(\boldsymbol{x},\boldsymbol{x}\right)=\left(\frac{1}{2}\boldsymbol{x}^{T}Q\boldsymbol{x}\right)\left(\frac{1}{2}\boldsymbol{x}^{T}R\boldsymbol{x}\right) \end{array}$$ does $f(x,y)$ has to be convex if it's given that $g(x)$ is not convex?

as follow to my other question here about the Gradient and Hessian of $g\left(\boldsymbol{x}\right)$ We know that $$\begin{array}{c} \nabla g\left(\boldsymbol{x}\right)=\left(\frac{1}{2}\boldsymbol{x}^{T}R\boldsymbol{x}\right)Q\boldsymbol{x}+\left(\frac{1}{2}\boldsymbol{x}^{T}Q\boldsymbol{x}\right)R\boldsymbol{x}\\ \nabla^{2}g\left(\boldsymbol{x}\right)=R\boldsymbol{x}\boldsymbol{x}^{T}Q+\left(\frac{1}{2}\boldsymbol{x}^{T}R\boldsymbol{x}\right)\cdot Q+Q\boldsymbol{x}\boldsymbol{x}^{T}R+\left(\frac{1}{2}\boldsymbol{x}^{T}Q\boldsymbol{x}\right)\cdot R \end{array}$$

I tried so far to calculate the hessian of $f(x,y)$ in oreder to check if the hessian is psd: it's became messy but I found it : $$\nabla^{2}f\left(\boldsymbol{x}\boldsymbol{y}\right)=\left(\begin{array}{cc} \frac{1}{2}\left(\boldsymbol{y}^{T}R\boldsymbol{y}\right)Q & \frac{1}{4}\left(Q\boldsymbol{x}\boldsymbol{y}^{T}R\right)\\ \frac{1}{4}\left(R\boldsymbol{y}\boldsymbol{x}^{T}Q\right) & \frac{1}{2}\left(\boldsymbol{x}^{T}Q\boldsymbol{x}\right)R \end{array}\right)\in\mathbb{R}^{2n\times2n}$$ but I dont really sure how to continue from here.(don't really sure if i can find if the hessian of $f$ is psd)

thanks!

Answer 1

I found the answer:$g$ is a special case of $f$ and therefore $f$ is not convex. proof: - $g$ not convex : $$A:\exists\boldsymbol{x}_{1},\boldsymbol{x}_{2}\in\mathbb{R}^{n},\exists t_{0}\in\left[0,1\right]:g\left(t\boldsymbol{x}_{1}+\left(1-t\right)\boldsymbol{x}_{2}\right)>tg\left(\boldsymbol{x}_{1}\right)+\left(1-t\right)g\left(\boldsymbol{x}_{2}\right)$$ - suppose $f$ was convex: then $$\begin{array}{c} \forall t\in\mathbb{R},t\in\left[0,1\right]:\\ \forall\left(\boldsymbol{w}_{1},\boldsymbol{w}_{2}\right),\left(\boldsymbol{z}_{1},\boldsymbol{z}_{2}\right)\in\text{dom}\left(f\right)=\mathbb{R}^{n}\times\mathbb{R}^{n}\Rightarrow f\left(t\boldsymbol{w}_{1}+\left(1-t\right)\boldsymbol{z}_{1},t\boldsymbol{w}_{2}+\left(1-t\right)\boldsymbol{z}_{2}\right)\leq tf\left(\boldsymbol{w}_{1},\boldsymbol{w}_{2}\right)+\left(1-t\right)f\left(\boldsymbol{z}_{1},\boldsymbol{z}_{2}\right) \end{array}$$ - for $$\begin{array}{c} \forall t\in\mathbb{R},t\in\left[0,1\right]:\\ \begin{array}{c} \left(\boldsymbol{w}_{1},\boldsymbol{w}_{2}\right)=\left(\boldsymbol{x}_{1},\boldsymbol{x}_{1}\right)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\\ \left(\boldsymbol{z}_{1},\boldsymbol{z}_{2}\right)=\left(\boldsymbol{x}_{2},\boldsymbol{x}_{2}\right)\in\mathbb{R}^{n}\times\mathbb{R}^{n} \end{array}\Rightarrow f\left(t\boldsymbol{x}_{1}+\left(1-t\right)\boldsymbol{x}_{2},t\boldsymbol{x}_{1}+\left(1-t\right)\boldsymbol{x}_{2}\right)\leq tf\left(\boldsymbol{x}_{1},\boldsymbol{x}_{1}\right)+\left(1-t\right)f\left(\boldsymbol{x}_{2},\boldsymbol{x}_{2}\right) \end{array}$$ - we know $\forall\boldsymbol{x}\in\mathbb{R}^{n}:g\left(\boldsymbol{x}\right)=f\left(\boldsymbol{x},\boldsymbol{x}\right)$ so we got $$\begin{array}{c} \forall t\in\mathbb{R},t\in\left[0,1\right]:\\ g\left(t\boldsymbol{x}_{1}+\left(1-t\right)\boldsymbol{x}_{2}\right)\leq tg\left(\boldsymbol{x}_{1}\right)+\left(1-t\right)g\left(\boldsymbol{x}_{2}\right) \end{array}$$ - Let $t=t_{0}$ we got $$B:g\left(t_{0}\boldsymbol{x}_{1}+\left(1-t_{0}\right)\boldsymbol{x}_{2}\right)\leq t_{0}g\left(\boldsymbol{x}_{1}\right)+\left(1-t_{0}\right)g\left(\boldsymbol{x}_{2}\right)$$ - we proof $A$ and $B$ and that's a contradiction. $\Rightarrow f$ is not convex.

Answer 2

A shorter argument: A function is said to be convex if it is convex along every line in the domain. $g$ is $f$ along the line $y=x$. Since, $g$ is not convex. $f$ cannot be convex.