Prove if the function is coercive

Question

Thank you for the comments and answers!

Show if the function is coercive: $$f(x,y) = x^2 + y^2 - xy - x$$

Have some difficulties dealing with the last $x$ term, tried to replace it with $|xy|$, didn't work. Could someone give me some hints of how to approach such question.

Note: definition of coercive, $$f(z)\rightarrow +\infty, as \ \|z\|\to\infty $$

Answer 1

Hint: $f(z)=z^THz+c^Tz$ with invertible $H$ can be written as $$ f(z)=(z+\frac12H^{-1}c)^TH(\underbrace{z+\frac12H^{-1}c}_{\hat z})-\frac14c^TH^{-1}c=g(\hat z). $$ Now $f$ is coercive iff $g$ is coercive.

Answer 2

Here is a purposely less refined answer.

You have a part $x^2 + y^2$ which looks coercive and a part $-xy - x$ which does not.

Let's borrow some of the coercive part to compensate $x y$ first: $$ |xy| \leq \tfrac12 (x^2 + y^2) , $$ so that $$ f(x,y) = x^2 + y^2 - xy - x \geq \tfrac12 (x^2 + y^2) - x . $$

The only way this could fail to be coercive is when $-x$ is very small; but then $\tfrac12 x^2$ is much larger than $|x|$, compensating again.

Answer 3

Using polar coordinates: $$ f(r,\theta)=r^2-r^2\cos\theta\sin\theta-r\cos\theta=r^2(1-\cos\theta\sin\theta)-r\cos\theta $$ In the direction $\theta=\pi/2$: $$ f(r,\pi/2)=r^2\rightarrow \infty\quad \mbox{when}\; r\rightarrow \infty $$