Definition. A continuous function $f(x)$, defined on $\mathbb{R^n}$, is coercive if: $$\lim_{||x||\xrightarrow[]{}\infty}f(x)=+\infty$$
Problem. Is the following solution is correct:
$f(x,y)=x^2-3xy+5y^2\\\hspace{1.3cm}
=x^2-3xy+(\frac{3}{2}y)^2+\frac{11}{4}y^2\\\hspace{1.3cm}
=(x-\frac{3}{2}y)^2+\frac{11}{4}y^2\\\hspace{1.3cm}
\geq\frac{11}{4}y^2 $
Since $\frac{11}{4}y^2\xrightarrow[]{}+\infty$, as $|y|\xrightarrow[]{}\infty$, can we conclude that $f(x,y)\xrightarrow[]{}+\infty$, as $||(x,y)||\xrightarrow[]{}\infty$ ?
If this solution is not correct: why exactly is it incorrect & what would be a correct approach?
If this solution is correct: why do we consider the function $\frac{11}{4}y^2$ on a plane rather than in space? Would its limit as $||(x,y)||\xrightarrow[]{}\infty$ be $+\infty$ in space? For example, if we move on the line $y=0$ in space, we get $\frac{11}{4}y^2=0$ for all values of $x$, instead of it infinitely increasing.
Best Answer
No, it is not correct because one can have sequence $(x_n,y_n)$ such that $\|(x_n,y_n)\| \to \infty$, but $|y_n|$ is bounded. For example, $(n,0)$.
However, you can also write $f(x,y) = \frac{11}{20}x^2+\left(y\sqrt{5}-\frac{3 x}{2 \sqrt{5}}\right)^2 \geq \frac{11}{20}x^2$. Together with $f(x,y)\geq \frac{11}4 y^2$, we have $$\frac{20}{11}f(x,y) + \frac 4{11}f(x,y)\geq x^2 + y^2 = \|(x,y)\|^2\\ \!\!\!\!\!\!\implies f(x,y)\geq \frac{11}{24}\|(x,y)\|^2\to\infty\text{ when } \|(x,y)\|\to\infty.$$
Note. I used Euclidean norm even though you don't specify it, but it works with any norm on $\mathbb R^2$ since all are equivalent.