I was looking through proofs of bounds of functions that didn't rely on calculus and one stackexchange topic I came across is this:

Show that $f(x,y)=\frac{xy^2}{x^2+y^4}$ is bounded

The goal is to show $|f(x,y)|$ is less than some real number. The quote on quote "answer", which is extremely presumptuous with almost no reasoning whatsoever, claims

"If $z>1$, then $z^2>z$." Okay, so what?

"If $z \leq 1$, then $z \leq 1+z^2$ since $z>0$." How does that follow and so what?

In math, you are expected to explain your reasoning. So, even if it is right, it's useless unless it can be argued why it makes sense that it is right. We are looking for the largest possible $f$, not of $z=x/y^2$.

  • $\begingroup$ After dividing both nominator and denominator by $y^4$ we obtain $$f(x,y)=\frac{z}{z^2+1}$$ for $z=x/y^2$. So all you need to study are different values of $z$. $\endgroup$ – Severin Schraven Jul 24 '19 at 7:01
  • $\begingroup$ And how do you know if you've studied "enough" of the values? How do you know you have maximized the output if for any two coordinates you plug in, you can always move infinitesimally further away to a potentially larger value? $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:04
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    $\begingroup$ Do not delete a question just to repost it a moment later. $\endgroup$ – Asaf Karagila Jul 24 '19 at 7:09
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    $\begingroup$ First: I think there is a mistake in either your copying or the original answer. In the second point it's supposed to be "since $z^2 > 0$". Second: I don't know if you have noticed, but the tone in your post is actually fairly presumptuous too. Please avoid that. Third: the answer explores the cases when $z > 1$ and when $z \le 1$. I Being $z$ real, I don't see a lot more cases to study. $\endgroup$ – Niki Di Giano Jul 24 '19 at 7:16
  • $\begingroup$ You don't really get to speak on behalf of other people. Your presumption that it's true doesn't mean that it is consistent to everyone else, and in fact intrinsically you are subconsciously relying on your trust in a mesh of observations that it is true. I trust that the answer is accurate because it wouldn't have been an accepted answer for this long without someone saying it is false if it was actually false, but the fact that I have to "trust" it instead following the steps they presumed but refused to show means that the answerer did not thoroughly explain their reasoning. $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:19

The point is that the function $g(z) = z/(1+z^2)$ satisfies $|g(z)| \le 1$ for all $z$. Since, as stated in the question, $f(x,y) = g(x/y^2)$, it must also be true that $|f(x,y)|\le 1$ for all $x,y$.

Note that the top answer is not a full proof of the fact, but rather filling in a gap in the question asker's proof.

  • $\begingroup$ Suppose you substitute "1" back into the original $f(x,y)$ function. That mere substitution does not show itself to be the largest possible output of the original function. So why then does $x/y^2$ being bounded imply the original function is bounded? $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:03
  • $\begingroup$ @cheesemonkey102 Since $f(x,y) = g(x/y^2)$, the range of $f$ must be contained in the range of $g$. $\endgroup$ – eyeballfrog Jul 24 '19 at 7:05
  • $\begingroup$ Okay, let's take that is true and use it. Then, how does it follow from the selected answer that answerer has found the maximum range? If they haven't found the maximum range, then that principal can't be used. They claimed bounds on two separate sub-expressions with different inequalities, which means you cannot carry operations across them like adding or subtracting the inequalities from each other. $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:07
  • $\begingroup$ @cheesemonkey102 The answer shows that if $|z| \ge 1$, then $z^2 > |z|$ (which implies $z^2 + 1 > |z|$), and if $|z| \le 1$, then $1+z^2 > |z|$. This exhausts all possiblities for $z$, showing $z^2 + 1 > |z|$ for all $z$, and thus $|g(z)| < 1$ for all $z$. $\endgroup$ – eyeballfrog Jul 24 '19 at 7:14
  • $\begingroup$ Alright, I think pieced together this essentially concludes that because the denominator is always greater than the numerator, the fraction of the two expressions is always less than one. Thank you. $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:23

The answer reasons as follows: we know that $f(x, y)$ can be written as $g(z)$ where $z = \frac{x}{y^2}$.

Now suppose $z > 1$. By multiplying this inequality by $z$ we get $z^2 > z$. Since $a + 1 > a$ for any $a \in \Bbb{R}$, $z^2 + 1 > z^2 > z$. Thus $1 > \frac{z}{1 + z^2}$.

Suppose $z \le 1$. Since $z^2 \ge 0$, it must be true that $z \le 1 \le 1 + z^2$. Thus $ 1 \ge \frac{z}{1 + z^2}$.

This exhausts all possibilities because $z \in \Bbb{R}$.

The function is symmetric (odd) which means that if we find an upper bound, this must also be a lower bound. If you need a proof of this, suppose we find an upper bound for an odd function unbounded from below (say, for example, that it goes to $-\infty$ for $z$ that goes to $\infty$). Since $g(-z) = - g(z)$, this means that $z \to - \infty \implies g(z) \to \infty$, which contradicts the assumption that $g$ is bounded above.

The answer is supposed to fill in the gaps of the question. I would avoid ranting about an incomplete proof because I couldn't fully understand it in the future.

  • $\begingroup$ I appreciate your work, but eyeball is who originally helped me to the answer and I selected their answer as correct considering the comments are publicly visible. $\endgroup$ – cheesemonkey102 Jul 24 '19 at 7:47
  • $\begingroup$ It's fine, it's the just decision. $\endgroup$ – Niki Di Giano Jul 24 '19 at 8:00

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