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I have attempted this question by using the fact that a composition of convex functions is also convex, and we know that $x^2+y^2$ is convex. However, I do not know how to show that $-axy$ is also convex. Do I need to prove it formally using the $\lambda \in [0,1]$ method?

Jochen
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ripbozo
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  • Use the definition of convexity hello? –  Oct 27 '22 at 09:14
  • Hi :) level sets of convex functions are convex. Can you prove, that $f(x, y) = x^2+y^2-axy$ is convex for $|a|\leq 2$? – Jochen Oct 27 '22 at 09:14
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    Can you prove the convexity of $x^2 + y^2 - axy$ by writing $$x^2 + y^2 - axy=\left(x-\frac a2y\right)^2+\frac{4-a^2}4y^2\ \ ?$$ Note that $4-a^2\geq0$. – Feng Oct 27 '22 at 09:15
  • @Jochen should i simply find whether the hessian of the function is positive semidefininite/definite ? – ripbozo Oct 27 '22 at 09:30
  • Yes, that's a good idea. – Jochen Oct 27 '22 at 10:21

1 Answers1

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$$x^2+y^2-axy=((1-a/2)(x+y)^2+(1+a/2)(x-y)^2)/2$$ $(x+y)^2$ and $(x-y)^2$ are both squares hence convex, so this is convex if $-2\le a\le 2$ and none of the coefficients are negative.

Proof that any square is convex follows from $(a-b)^2\geq0$ thus $$a^2+b^2\geq2ab\tag{1}$$ so $$ \begin{align} ((1-t)a+tb)^2 &=(1-t)^2a^2+2t(1-t)ab+t^2b^2\\ &\geq(1-t)^2a^2+t(1-t)(a^2+b^2)+t^2b^2\\ &=(1-t)a^2+tb^2\\ \end{align} $$ The proof that the sum of two convex function is convex is very obvious.

Suzu Hirose
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