How can I use the concept of a proper map to show that a set is bounded?

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

cbarker1 said:

Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \\ x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

What about using ##(x -2y)^2 = x^2 -4xy + 4y^2##?

Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.

cbarker1 said:

Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.

Why not rewrite the original inequality using that equation?

cbarker1 said:

Now, how I do use an inequality to eliminate the ##(x-2y)^2##? At the end, I just want this fact ##x^2+y^2\leq 5##.

PS in case you missed the point that ##(x-2y)^2 \ge 0##

cbarker1 said:

Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

Complete the square in one of the variables:

[tex]3x^2 - 4xy + 5y^2 = 3\left(x - \frac23y\right)^2 + \left(5 - \frac 43\right)y^2[/tex]

PeroK said:

PS in case you missed the point that ##(x-2y)^2 \ge 0##

I have been working on this problem for a couple of weeks. So I did not see that helpful fact. Thanks.

That squares are non-negative is generally one of the most widely used tricks in mathematics!

cbarker1 said:

Summary: Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:
##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##
##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.
Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks
cbarker1

This is the answer:
##3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2\leq x^2+(x-2y)^2+y^2##
Since ##(x-2y)^2\geq 0## then ##x^2+y^2\leq 5 \implies \sqrt{x^2+y^2}\leq \sqrt{5}##.

cbarker1 said:

This is the answer:
##3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2\leq x^2+(x-2y)^2+y^2##
Since ##(x-2y)^2\geq 0## then ##x^2+y^2\leq 5 \implies \sqrt{x^2+y^2}\leq \sqrt{5}##.

I don't follow this at all. Instead:
$$3x^2-4xy+5y^2 =2x^2+(x-2y)^2+y^2 \ge x^2+ y^2$$

Another option is to set [itex](x,y) = r(\cos\theta, \sin\theta)[/itex] and show that the inequality can be written as [tex]r^2 \leq \frac{5}{A - B\cos(2(\theta-\theta_0))}[/tex] for some [itex]\theta_0[/itex] where [itex]A > B > 0[/itex]. Then [tex]
x^2 + y^2 = r^2 \leq \frac{5}{A - B}.[/tex]

another soution, similar to others: the given expression equals 2(x-y)^2 + x^2 + 3y^2. but

x^2+y^2 ≤ 2(x-y)^2 + x^2 + 3y^2 ≤ 5, since squares are non negative.

In general, as I recall, there are basically 3 types of quadratic expressions in (at most) 2 variables, x^2 = 1, x^2+y^2 = 1, and xy = 1, all others being essentially equivalent to one of these, only one of which is bounded in the xy plane. So you have to detect which type is yours. or you may consider -x^2 - y^2 and x^2-y^2 as somewhat different, but since x^2-y^2 = (x-y)(x+y) = uv, and -x^2-y^2 = -(x^2+y^2), you see these are closely related to the first 3.

Now essentially you want to measure how much your example is dominated by the terms x^2 and y^2, or by the term xy, so from ax^2 + 2bxy + cy^2, you form ac-b^2. Then your example is bounded, i.e. looks like x^2 + y^2 = 1, if and only if this expression is positive. In your case a= 5, c = 3, and 2b = 4, so ac-b^2 = 11 > 0, and it is bounded.

The reason is that one can "diagonalize" any example so that it looks like ax^2 + cy^2, where a and c can be zero or positive or negative. This is done by writing the quadratic form as a symmetric matrix expression, and diagonalizing that matrix. (This is equivalent to completing the square, i.e. getting rid of the coefficient of xy, as discussed above).

In your case the matrix has rows [5 -2] and [-2 3], and the determinant, ac-b^2 = 11. When your matrix is diagonalized, the entries on the diagonal, the new a and c, will be the roots of the characteristic polynomial of this (original) matrix, and their product will equal the determinant of this (original) matrix. So if that determinant is positive, then the roots will have the same sign and your quadratic will be equivalent to either x^2+y^2 or -(x^2+y^2), and both x^2+y^2 = 1 and -(x^2+y^2) = 1, are bounded, (since empty implies bounded).

To summarize, unless your quadratic form ax^2 + 2bxy + cy^2 actually defines a (doubled) straight line, or the empty set, it is equivalent to either x^2 + y^2, or x^2 - y^2, and you can tell which by computing ac-b^2. Indeed you only need the sign, which may be much easier than the number itself. E.g. if the coefficients a and c have opposite sign, you know ac-b^2 is negative, as in the case of say 3789x^2 - 5613 xy - 8201 y^2.

A different approach is to do a translation/rotation of the right type, to make the xy-term disappear. of course, this will still require additional work, but it can give you insights. This case looks like you have a rotated ellipse. Rotations are isometries, meaning they preserve the distances between points, so set is bounded iff its rotated image is bounded. Basically, your new coordinates (x',y') will be rotated images of (x,y) , and then you set things so that the term x'y' is 0. This is a similar idea to that in mathwonk's post #13 above, right before this one.

Indeed the nice discussion in #14 gives a geometric way to view the algebraic "diagonalization" process referenced in # 13. I.e. to diagonalize, one finds the 2 eigenvalues and 2 corresponding eigenvectors of the symmetric matrix. It follows that the two eigenvectors are perpendicular, so the change of basis matrix they define (the matrix that diagonalizes the original one) can be chosen to be a rotation (take eigenvectors of length one and in the appropriate order).

nuuskur said:

Note that
[tex]
3x^2 - 4xy + 5y^2 = 2(x-y)^2 + x^2 +3y^2 \geqslant 0.
[/tex]
Put ##f(x,y)=3x^2-4xy+5y^2##. One has ##S=f^{-1}([0,5])##. Preimage of closed and bounded (i.e compact) under continuous map is also closed and bounded. Polynomials are smooth (hence continuous).

The domain of ##f## is not compact.

nuuskur said:

Preimage of closed and bounded (i.e compact) under continuous map is also closed and bounded. Polynomials are smooth (hence continuous).

Take ##f(x)=5## for ##x\in\mathbb R##, it is continuous but the preimage of the closed and bounded set ##\{5\}## is the whole of ##\mathbb R## which is not bounded.

Oopsies. My bad.

this is an easy mistake. the concept of a map for which the preimage of a compact set is also compact is very important in algebraic geometry and called a "proper map". if X and Y are compact Hausdorff spaces, then all continuous maps X-->Y are proper (and closed).

On metric spaces proper is equivalent to being a closed map with the inverse image of every point compact, neither of which is generally true for all continuous maps.

the definition of proper used in algebraic geometry is in terms of the forward properties of maps and emphasizes the fact that for say locally compact hausdorff spaces, proper is equivalent to being "universally closed", i.e. not only is X->Y closed, but so is XxZ-->YxZ for all Z.

Many of us have made the slip of failing to assume properness to derive some nice property of maps, including (me and) even the famous algebraic geometer I. Shafarevich, whose argument for the corollary of theorem 7 p. 61 (upper semi continuity, on the base, of fiber dimension) in his wonderful book Basic Algebraic Geometry, 1974 edition, is lacunary for this reason.