MHB Is Set M1 Convex? A Proof Using Mathematical Induction

[mathmari]

Hey!

We have the function $\displaystyle{y=f(x_1, x_2)=x_1\cdot x_2^2}$ and the set $M_1=\{x\in [0, \infty)^2 \mid f(x_1, x_2)>1\}$.

I want to check if the set is convex. Let $x=(x_1, x_2) , y=(y_1, y_2)\in M_1$, then $x_1\cdot x_2^2>1$ and $y_1\cdot y_2^2>1$.

We want to show that \begin{equation*}\lambda x+(1-\lambda )y=\lambda (x_1, x_2)+(1-\lambda )(y_1, y_2)=(\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2)\in M_1\end{equation*} so we have to show that \begin{equation*}f\left (\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2\right )>1\end{equation*}

We have the following:
\begin{align*}f&\left (\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2\right )=(\lambda x_1+(1-\lambda )y_1)\cdot( \lambda x_2 +(1-\lambda )y_2)^2 \\ &=(\lambda x_1+(1-\lambda )y_1)\cdot( \lambda^2 x_2^2+2\lambda x_2(1-\lambda )y_2 +(1-\lambda )^2y_2^2)\\ &= \lambda x_1\cdot( \lambda^2 x_2^2+2\lambda (1-\lambda )x_2y_2 +(1-\lambda )^2y_2^2)+(1-\lambda )y_1\cdot( \lambda^2 x_2^2+2\lambda (1-\lambda )x_2y_2 +(1-\lambda )^2y_2^2) \\ & = \lambda^3 x_1x_2^2+2\lambda^2 (1-\lambda )x_1 x_2y_2 +\lambda (1-\lambda )^2x_1 y_2^2+ \lambda^2 (1-\lambda )x_2^2y_1+2\lambda (1-\lambda )^2x_2y_1y_2 +(1-\lambda )^3y_1y_2^2 \\ & > \lambda^3 +2\lambda^2 (1-\lambda )x_1 x_2y_2 +\lambda (1-\lambda )^2x_1 y_2^2+ \lambda^2 (1-\lambda )x_2^2y_1+2\lambda (1-\lambda )^2x_2y_1y_2 +(1-\lambda )^3\end{align*}

Is this correct so far? (Wondering)

How could we continue? (Wondering)

 

[Opalg]

We have the function $\displaystyle{y=f(x_1, x_2)=x_1\cdot x_2^2}$ and the set $M_1=\{x\in [0, \infty)^2 \mid f(x_1, x_2)>1\}$.
I want to check if the set is convex.
Let $x=(x_1, x_2) , y=(y_1, y_2)\in M_1$, then $x_1\cdot x_2^2>1$ and $y_1\cdot y_2^2>1$.
We want to show that \begin{equation*}\lambda x+(1-\lambda )y=\lambda (x_1, x_2)+(1-\lambda )(y_1, y_2)=(\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2)\in M_1\end{equation*} so we have to show that \begin{equation*}f\left (\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2\right )>1\end{equation*}
We have the following:
\begin{align*}f&\left (\lambda x_1+(1-\lambda )y_1, \lambda x_2 +(1-\lambda )y_2\right )=(\lambda x_1+(1-\lambda )y_1)\cdot( \lambda x_2 +(1-\lambda )y_2)^2 \\ &=(\lambda x_1+(1-\lambda )y_1)\cdot( \lambda^2 x_2^2+2\lambda x_2(1-\lambda )y_2 +(1-\lambda )^2y_2^2)\\ &= \lambda x_1\cdot( \lambda^2 x_2^2+2\lambda (1-\lambda )x_2y_2 +(1-\lambda )^2y_2^2)+(1-\lambda )y_1\cdot( \lambda^2 x_2^2+2\lambda (1-\lambda )x_2y_2 +(1-\lambda )^2y_2^2) \\ & = \lambda^3 x_1x_2^2+2\lambda^2 (1-\lambda )x_1 x_2y_2 +\lambda (1-\lambda )^2x_1 y_2^2+ \lambda^2 (1-\lambda )x_2^2y_1+2\lambda (1-\lambda )^2x_2y_1y_2 +(1-\lambda )^3y_1y_2^2 \\ & > \lambda^3 +2\lambda^2 (1-\lambda )x_1 x_2y_2 +\lambda (1-\lambda )^2x_1 y_2^2+ \lambda^2 (1-\lambda )x_2^2y_1+2\lambda (1-\lambda )^2x_2y_1y_2 +(1-\lambda )^3\end{align*}

Is this correct so far? (Wondering)

It may be correct, but it's too complicated to be helpful. I would be surprised if you could solve this problem by working directly from the definition of convexity.

The condition for $(x,y)$ to be in $M_1$ is $xy^2>1.$ If you write that as $y > \dfrac1{\sqrt x}$, it says that $(x,y)$ lies above the graph of the function $y = \dfrac1{\sqrt x}$. But that function is a convex function (because its second derivative is positive), and there is then a theorem to say that the region above its graph is a convex set.

 

[mathmari]

I got it! (Nerd)

Thank you very much! (Smile)