Showing coercivity of a function

Question

I am well attuned to the definition for a function to be coerce, which is that $\lim_{\|x\| \to \infty}f(x) = \infty$ ie the values of $f$ go to infinity as the norm goes to infinity.

So

Ex.1 $f(x_1,x_2) = x_1^4 + x_2^4 - 3x_1x_2$

I thought this was immediately obvious because $x_1^4$ and $x_2^4$ grow faster than the linear terms we have. Would this be considered a sufficient proof?

I am unsure in particular how taking the limit would be shown.

Additionally, I don't see why functions like

Ex.2 $f:\mathbb{R}^n \to \mathbb{R}, f(x) = a^tx$ isn't coercive,

or

Ex.3 $f(x_1,x_2) = x_1^2 + x_2^2 - 2x_1x_2$ isn't coercive.

Since they are both unbounded, doesn't taking the norm of $x \to \infty$ show that these functions also approach infinity?

Answer 1

Ex.1 No, your answer is not rigorous. It is true, but you need to prove it. My suggestion is to show that $$\lim_{\|(x_1,x_2)\| \to +\infty} \frac{x_1 x_2}{x_1^4+x_2^4}=0.$$

Ex.2 If $x \perp a$, then $f(x)=0$. And since on the subspace $\{a\}^\perp$ there are vectors of arbitrarily large norm...

Ex. 3 If $x_1=x_2=t$, then $f(t,t)=0$ for any $t>0$. Letting $t \to +\infty$...

Answer 2

$$f(x_1, x_2) = x_1^2 + x_2^2 - 2x_1x_2$$ is not coercive because $f(x_1,x_2) = (x_1 - x_2)^2$, meaning that $f(x,x) = 0$ no matter how large $x$ is.

This means that the statement

For all $M>0$, there exists a $N>0$ such that for all $x, ||x||> N$, it is true that $f(x) > M$

Is not true for this function.

Similarly, for $f(x) = a^T x$, you know that if $v$ is perpendicular to $a$, then $f(\alpha v) = 0$ for every value of $\alpha \in \mathbb R$.