For what conditions on a,b,c is both $\frac{x^2+y^2}{|ax^2+bxy+cy^2|}$ and $\frac{|ax^2+bxy+cy^2|}{x^2+y^2}$ bounded?

Question

I'm not sure how exactly to approach this problem, here is what I've done so far:

I've shown that $$\frac{|ax^2+bxy+cy^2|}{x^2+y^2}= a\frac{x^2}{x^2+y^2}+b\frac{|xy|}{x^2+y^2}+c\frac{y^2}{x^2+y^2} \leq a+b+c$$ Meaning there are no restrictions on the polynomial coefficients for this fraction. It remains then to prove the reciprocal is bounded above, which is what I'm struggling with: $$\exists M, \frac{x^2+y^2}{|ax^2+bxy+cy^2|}<M$$

I had two ideas so far. Since the level curve of the numerator is a circle and the level curve of the denominator is an ellipse with rotated axes, I was thinking of defining some kind of elliptical norm $[x,y]=|ax^2+bxy+cy^2|$ and showing that $[x,y] \geq |(x,y)|$ under certain conditions. However, I'm not sure how to show this and how it would even lead me to my desired result.

The second idea was that showing that if the denominator never reaches $0$, ie. the discriminant $< 0$ then the expression never gets arbitrarily large. This leads to the inequality $b^2<4ac$, however I'm once again not sure how I would go about proving this is actually the condition I need.

Am I on the right track for either of these? If so, how should I go about formally showing they are correct?

For second-year analysis course.

Answer 1

Let $(x,y)=(r\cos \theta,r\sin \theta)$, then

\begin{align} f(x,y) &= \frac{ax^2+bxy+cy^2}{x^2+y^2} \\ &= a\cos^2 \theta+b\cos \theta \sin \theta+c\sin^2 \theta \\ &= \frac{a(1+\cos 2\theta)}{2}+\frac{b\sin 2\theta}{2}+\frac{c(1-\cos 2\theta)}{2} \\ &= \frac{a+c}{2}+\frac{a-c}{2} \cos 2\theta+\frac{b}{2} \sin 2\theta \\ \end{align}

Now $$\frac{a+c-\sqrt{(a-c)^2+b^2}}{2} \le f(x,y) \le \frac{a+c+\sqrt{(a-c)^2+b^2}}{2}$$

Imposing removable singularities for $f(x,y)$ and $\dfrac{1}{f(x,y)}$ at $(x,y)=(0,0)$, we have

$$0\le \sqrt{(a-c)^2+b^2} < |a+c|$$

which means $$a+c \ne 0 \quad \text{and} \quad b^2-4ac<0$$

Please also refer to another questions here and here.

Answer 2

I assume $a,b,c\in \mathbb R.$ As you indicated, the easy direction is

$$\frac{|ax^2+bxy+cy^2|}{x^2+y^2} \le \frac{|a|x^2+|b||xy|+|c|y^2|}{x^2+y^2} \le |a| + |b|/2 +|c|.$$

So that fraction is bounded on $\mathbb R^2 \setminus \{(0,0)\}$ for all $a,b,c.$

For the reciprocal, let $p(x,y) = ax^2+bxy+cy^2.$ Claim: $(x^2+y^2)/|p(x,y)|$ is bounded on $\mathbb R^2\setminus \{(0,0)\}$ iff $p$ never vanishes on $\mathbb R^2\setminus \{(0,0)\}.$ Proof: If the expression is bounded, then clearly $p(x,y)\ne 0$ for all $(x,y)\in \mathbb R^2\setminus \{(0,0)\}.$ In the other direction, let $C$ be the unit circle. Since $p$ is nonzero on $C,$ continuity shows $\min_C |p| >0.$ From this and the second order homogeneity of numerator and denominator, we get

$$\frac{x^2+y^2}{|p(x,y)|}\le \frac{1}{\min_C |p|}\,\, \text {for all } (x,y) \in \mathbb R^2\setminus \{(0,0)\}.$$

So when is $p(x,y)$ nonvanishing on $\mathbb R^2\{(0,0)\}?$ It turns out that this is equivalent to $b^2-4ac<0.$ One way to check this is to consider the lines $y=mx$ for $m\in \mathbb R.$ I'll leave this to you for now; feel free to ask about it.

Thus the answer to the question in the title is: Iff $b^2-4ac<0.$