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There is an exercise in my lists about those functions: $$f(x, y) = (y-3x^2)(y-x^2) = 3 x^4-4 x^2 y+y^2$$

$$g(t) = f(vt) = f(at, bt); a, b \in \mathbf{R}$$

It asks me to prove that $t = 0$ is a local minimum of $g$ for all $a, b \in \mathbf{R}$

I did it easily: $$g(t) = 3 a^4 t^4-4 a^2 t^2 b t+b^2 t^2$$ $$g'(t) = 2 b^2 t-12 a^2 b t^2+12 a^4 t^3$$ $$g''(t) = 2 b^2-24 a^2 b t+36 a^4 t^2$$

It is a critical point: $$g'(0) = 0; \forall a, b$$

Its increasing for all a, b: $$g''(0) = 2b^2 > 0; \forall b \ne 0$$ and $$b = 0 \implies g(t) = 3 a^4 t^4$$ which has only one minimum, at $0$, and no maximum

However, it also asks me to prove that $(0, 0)$ is not a local minimum of $f$. How can this be possible? I mean, if $(0, 0)$ is a minimum over every straight line that passes through it, then, in this point, $f$ should be increasing in all directions, no?

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You've tried plotting your function? – J. M. Oct 19 '11 at 7:46
Approach $(0,0)$ along the parametric path $x=t$, $y=2t^2$. – André Nicolas Oct 19 '11 at 7:51
Yes, and this seems inconsistent with what I found about $g$ – rogi Oct 19 '11 at 7:52
Even a simple-looking function of two variables can do weird things. – André Nicolas Oct 19 '11 at 7:55
It is zero on the curves $y=x^2$ and $y=3x^2$. It is negative when $x^2 < y < 3x^2$. – AD. Oct 19 '11 at 7:58
up vote 6 down vote accepted

Using the factorisation you have we note that $f$ is:

  • zero on the curves $y=x^2$ and $y=3x^2$
  • negative when $x^2<y<3x^2$
  • positive otherwise.

Since there are points arbitrary close to $o=(0,0)$ that satisfy $x^2<y<3x^2$ (say $y=2x^2$ where $x$ is small) and points arbitrary close to $o$ that do not satisfy $x^2<y<3x^2$ (say $y=0$ and small $x$) we see that $o$ is not a local minimum.

EDIT Note that in any neighbourhood of $o$ there is no line segment through $o$ which is contained in the region $x^2<y<3x^2$.

2nd Edit Draw a line through $o$, then there is a line segment surrounding $o$ such that the line segment has no point $(x,y)$ such that $x^2<y<3x^2$. I hope this is sufficiently clear - otherwise just ask!

Descriptive picture


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Ok, but if there is an "arbitrarily close" to $(0, 0)$ that is negative, can't we trace a straight line through it and teh origin? – rogi Oct 19 '11 at 8:20
Pick a definite point $(a,b)$, very very close to (0,0)$, at which the function is negative. Join $(a,b)$ to the origin by a line. By what you proved, as you approach the origin along that line, after a while the function will be positive. However, that does not change the fact that, to be fancy, for any $\epsilon>0$, there is a point $(a_\epsilon,b_\epsilon)$, at distance $<\epsilon$ from the origin, at which the function is negative. – André Nicolas Oct 19 '11 at 8:36
What I don't get is this "after a while teh function will be positive". It should be positive immediately, otherwise, $(0, 0)$ would not be minimum of $g$ – rogi Oct 19 '11 at 17:41
@AYGHOR What do you mean? – AD. Oct 19 '11 at 18:10
What I mean is: $P$ is a point as close as it can be to $(0, 0)$. Then there is a line from $(0, 0)$ to $P$. If $f(P)$ is negative, then $(0, 0)$ is not a minimum of $f$ restricted to this line. EDIT: I mean straight lines. – rogi Oct 19 '11 at 19:17

Saying that "$(0,0)$ is a local minimum of $f$ restricted to any line through the origin" means that, for every such line, there might be some bad points on the line at which $f$ takes negative values, but none of them are "close to the origin". Note that "close to the origin" can depend on the line we're talking about.

However, there are lots of lines through the origin, and each one can have some bad points. It's quite possible for the bad points on all the lines together to approach the origin, which can make $(0,0)$ not a local minimum of $f$.

That's the idea how the seeming paradox can be resolved. As for proving that it really happens: can you characterize the points $(x,y)$ at which $f(x,y) < f(0,0) = 0$? Are those points staying far away from $(0,0)$, or are they all around the origin?

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Sorry, that "bad point" idea did not make any sense to me – rogi Oct 19 '11 at 8:04
If "$f$ restricted to every straight line passing through teh origin" has a minimum at $(0, 0)$, then there should be at least circle, even if ridiculously small, around $(0, 0)$ over which teh values of $f$ are $> 0$, no? – rogi Oct 19 '11 at 8:14
@AYGHOR: No, that's not true. (This particular function $f$ is a counterexample!) – Hans Lundmark Oct 19 '11 at 9:27
This is the problem. Such a circle do not exists for your function. – xen Oct 19 '11 at 9:28
OMG I'M BLIND I CAN'T SEE WHY O_O – rogi Oct 19 '11 at 17:43

Draw the set of points in the $xy$-plane where $f(x,y) = 0$. Then look at the regions of the plane that are created and figure out in which of them $f(x,y)$ is positive and in which $f(x,y)$ is negative. From there, you should be able to prove that $(0,0)$ is neither a local maximum nor a local minimum point. (Hint: Using your sketch, is there any disk centered at $(0,0)$ in which $f(x,y)$ takes its minimum value at $(0,0)$?)

As important as understanding why the phenomenon you are observing is happening (no local minimum at the origin even though the function restricted to any straight line through the origin has a local minimum there) is to figure out how you would construct such a function and why the example given is, in some sense, the simplest kind of example you could construct. The idea used here is very similar to creating an example which shows that $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ may not exist even if the limit exists along all straight lines approaching the origin.

What you're really learning in this example is that the seemingly reasonable intuition that we would naturally have that we can understand a function of two variables near a point by understanding all of the functions of one variable obtained by restricting the function to lines through that point is misguided -- in particular, it fails if we are trying to find local maxima and minima.

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