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Just to talk about something fresh for which I still have a good memory of what I actually thought and what I wrote, let's take this examplethis example

Just to talk about something fresh for which I still have a good memory of what I actually thought and what I wrote, let's take this example

Just to talk about something fresh for which I still have a good memory of what I actually thought and what I wrote, let's take this example

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fedja
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Just to talk about something fresh for which I still have a good memory of what I actually thought and what I wrote, let's take this example

Thoughts:

  1. Ivan Fesenko teased me with the puzzle without the gradient condition 10 years ago. I solved it with the standard $(1-xy)^2+x^2$, but it would be nice to tease him with the upgraded puzzle when I talk to him next. Also, it is high time to finish it off.

  2. The standard polynomial is even and the only critical point is a saddle. If you think of it, this saddle is inevitable: we have low points on the landscape on both sides of the $x$-axis and high values on the $x$-axis going to $+\infty$ both ways. Thus the mountain pass lemma will ensure a saddle somewhere.

  3. This works every time when the sequence of points where the polynomial goes to $0$ has two different limiting directions. Then we can separate them by a line and run the same argument. So, the limiting direction must be unique.

  4. This seems impossible because the highest (even) degree homogeneous part should vanish in this direction but then it also vanishes in the opposite direction. This makes hte second highest (odd) degree homogeneous part to vanish on the entire line too. The next degree is unclear though...

  5. We certainly need something to break the symmetry here. A polynomial family $P_y(x)$ of polynomials in $x$ that have roots close to $0$ when $y\to+\infty$ and no roots when $y\to-\infty$ would be nice.

  6. Hey, I know this one: $yx^2-1$. Let's try $x^2+(x^2y-1)^2$.

  7. Damn, it doesn't work. The origin is still a critical point.

  8. Yeah, what else would you expect: the low points on the landscape are accumulating to one direction but still are separated by the line, so the mountain pass lemma is as powerful as before. To kill it, we need to shift both descends to one side.

  9. Add $x$ to $P_y(x)$. That won't change the limiting direction but will shift the zeroes a bit. So, let's try $f(x,y)=x^2+(x^2y+x+1)^2$.

  10. The origin is good now: $f_x=2$ there. Actually, it is $2$ everywhere where $x=0$.

  11. If $x\ne 0$, then $f_y=0$ only if $x^2y+x+1=0$ but then $f_x=2x\ne 0$ by the chain rule.

  12. OK, let's post the example and keep it in memory for teasing people...

That's what I actually thought for the last hour or so (interlaced with some personal thoughts that are of no interest for this thread).

What I wrote can be seen easily if you follow the link.

Why such discrepancy?

a) Some steps in the chain like 1) and 10) are too personal to be of interest to anybody. You need them to "start the engine running" and to "vent the steam", but they aren't, strictly speaking, mathematics and do not make me look any better, so why to publish them?

b) Some steps like 8) and the heuristics in 9) are actually false. To publish them would be ridiculous.

c) 4) and 7) are "failures" on the way. There is no point in telling anyone where and how I failed. I could fill volumes with my failed attempts if I started doing it.

d) 10) and 11) are trivial computations. Everyone can do those himself.

e) 2), 3), 5) are left. 9) is the counterexample. 2), 3) are steps in the direction of the affirmative answer. Once the final answer is negative, there is no point in talking of the steps in the opposite direction.

f) 5) is a nice idea but everybody knows that $y$ is not even. One can see the whole mechanics in the answer itself, so there is no need to explain it separately.

I don't know if this account of one personal affair with one relatively simple problem can really shed much light on why we do not tell/write exactly what we see/think of, but you asked and I answered.