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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?2011-10-19
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    Approach $(0,0)$ along the parametric path $x=t$, $y=2t^2$.2011-10-19
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    Yes, and this seems inconsistent with what I found about $g$2011-10-19
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    Or checking the endpoints?2011-10-19
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    Even a simple-looking function of two variables can do weird things.2011-10-19
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    What I mean is: if $t = 0$ is a minimum for $g$, then, every point "around" $(0, 0)$ must be higher than 0. Am I wrong?2011-10-19
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    It is zero on the curves $y=x^2$ and $y=3x^2$. It is negative when $x^2 < y < 3x^2$.2011-10-19

3 Answers 3

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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
  • positive otherwise.

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

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. I hope this is sufficiently clear - otherwise just ask!

Descriptive picture

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?2011-10-19
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    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.2011-10-19
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    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$2011-10-19
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    @AYGHOR What do you mean?2011-10-19
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    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.2011-10-19
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    @AYGHOR Did you see the 2nd Edit? No matter where you choose $P$ the line will always have a segment at $o$ which is not in the region where $f$ is negative. This is why $g$ has a local minimum there but not $f$.2011-10-19
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    @AD. after looking at this: http://www.wolframalpha.com/input/?i=plot+%28y-3x%5E2%29%28y-x%5E2%29+%3D+0 I think I do understand, but it just seem impossible.2011-10-19
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    @AYGHOR See it better now?2011-10-20
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    Is there any way this technique can be salvaged? For instance, if $g''$ had been positive at all $b$ (including $b=0$), would that have been sufficient for $0$ to be a local minimum? I would think so, since then the Hessian of $f$ must be positive definite (instead of semidefinite, as in this example) and so $f$ convex. Is there anything more general that can be said?2011-10-20
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    @user7530 If $g$ did not have a local min then $f$ would not have it either. However, I do not think the affirmative direction can is very useful in this case - for this family of $g$'s. Of course if instead of strait lines we looked at all curves then we would reach the goal - but that is not very useful I believe.2011-10-20
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    @AD. I did draw this same graphic and managed to see it yesterday. Thank you very much anyway, this show pretty well the situation! However, it seems just _wrong_ teh fact that all $g$ having minimum at $(0, 0)$ do not guarantee a circle of positive values.2011-10-21
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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 me2011-10-19
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    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?2011-10-19
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    @AYGHOR: No, that's not true. (This particular function $f$ is a counterexample!)2011-10-19
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    This is the problem. Such a circle do not exists for your function.2011-10-19
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    OMG I'M BLIND I CAN'T SEE WHY O_O2011-10-19
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    See sweetjazz's answer which addresses this very point.2011-10-20
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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.