Undergrad Understanding Cauchy-Schwarz Inequality

[Mr Davis 97]

I am trying to find the max and min values of the function $f(x,y) = 2\sin x \sin y + 3\sin x \cos y + 6 \cos x$. By the Cauchy-Schwarz inequality, we have that $|f(x,y)|^2 \le (4+9+36) (\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 x) = 49$. Hence $-7 \le f(x,y) \le 7$.

My question has to do with the last inequality. What information does this inequality convey exactly? Does it say that f(x,y) actually takes on all values and only the values in the interval $[-7,7]$,and hence -7 and 7 are the min and max? Or does it say -7 and 7 are bounds on the range of f(x,y), and the max and min could actually be smaller values contained in the interval?

 

[StoneTemplePython]

In general you are just bounding the maximum and minimum values here, and bound is loose unless you can find a case where one thing is a scalar multiple of another, i.e.

$\big \vert \langle \mathbf a, \mathbf b \rangle \big \vert \leq \big \vert \mathbf a\big \vert_2 \vert \mathbf b\big \vert_2$,

with equality iff $\mathbf a \propto \mathbf b$

(and there is the special case of one of the vectors being the zero vector).
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It isn't clear to me why you have

$(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1$

I.e. from what you've given $x$ and $y$ are independent variables. If there is some constraint that x = y, then it holds. Otherwise I can easily come up with a case where it equals 2.

 

[Mr Davis 97]

In general you are just bounding the maximum and minimum values here, and bound is loose unless you can find a case where one thing is a scalar multiple of another, i.e. $\big \vert \langle \mathbf a, \mathbf b \rangle \big \vert \leq \big \vert \mathbf a\big \vert_2 \vert \mathbf b\big \vert_2$, with equality iff $\mathbf a \propto \mathbf b$ (and there is the special case of one of the vectors being the zero vector).
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It isn't clear to me why you have

$(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1$

I.e. from what you've given $x$ and $y$ are independent variables. If there is some constraint that x = y, then it holds. Otherwise I can easily come up with a case where it equals 2.

I meant for that y to be an x, so it is 1.

But does this mean that -7 and 7 are not the min and the max?

 

[StoneTemplePython]

I meant for that y to be an x, so it is 1.

I don't know what this means. You explicitly defined a two variable function $f(x,y)$. But I'll assume the identity holds and move on:

But does this mean that -7 and 7 are not the min and the max?

$\mathbf a = \begin{bmatrix} 2\\ 3\\ 6 \end{bmatrix}$

$\mathbf b = \begin{bmatrix} \sin x \sin y\\ \sin x \cos y \\ \cos x \end{bmatrix}$

is there some $\gamma$, where $\mathbf a = \gamma \mathbf b$?

If not then Cauchy's inequality is strict and you can be certain the function never takes on the values of -7 and 7.

 

[FactChecker]

It isn't clear to me why you have

$(\sin^2 x \sin^2y + \sin^2 x \cos^2 y + \cos^2 y) = \sin^2(x) + \cos^2(y) = 1$

Especially since the OP has concluded that |f|≤7 but f(π/2, 0) = 9.

 

[FactChecker]

I meant for that y to be an x, so it is 1.

Do you mean in the original problem or just in this part? f(π/2, 0) = 9, so if you need to correct the original problem, you should do that and see what changes.

 

[FactChecker]

$\mathbf a = \begin{bmatrix} 2\\ 3\\ 6 \end{bmatrix}$

$\mathbf b = \begin{bmatrix} \sin x \sin y\\ \sin x \cos y \\ \cos y \end{bmatrix}$

is there some $\gamma$, where $\mathbf a = \gamma \mathbf b$?

If not then Cauchy's inequality is strict and you can be certain the function never takes on the values of -7 and 7.

It's interesting to note that at about x=0.550818287203578, the function f(x,x) is about maximized at about 7 and the vectors you give above are essentially multiples of each other.

 

[StoneTemplePython]

It's interesting to note that at about x=0.550818287203578, the function f(x,x) is about maximized at about 7 and the vectors you give above are essentially multiples of each other.

yea -- by my calculations

- - - -

edit:

we have

$\gamma \cos(x) = 6$
$6 \sin(x) = \big(\gamma \cos(x)\big) \sin(x) =\gamma \sin(x) \cos(x) = 3$
hence we have $\sin(x) = \frac{1}{2}$

Finally

$\gamma \frac{1}{2}\frac{1}{2} = \gamma \sin(x) \sin(x) = 2$
hence $\gamma = 8$

unfortunately this implies

$8 \cos(x) = 6$, or $\cos(x) = 0.75$, while in fact we have something more like $\cos(x) = 0.87$, so the $\gamma$ doesn't quite exist and the inequality is strict. Nevertheless, interesting to get so close to 7.