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Let $(x,y)\in[0,1)\times[0,1)$ cum $x^2+y^2<1$. Are there any $\mu\geq\lambda>0$ such that


$\forall\xi=(\xi_1,\xi_2)^T\in\mathbb R^2$ ?

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closed as off-topic by Did, Ahaan S. Rungta, Tim Raczkowski, Solid Snake, Davide Giraudo Jul 23 '15 at 18:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

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This sounds like it might be homework --- if it is, can you add the homework tag? – John Engbers May 25 '12 at 18:33
And where did you get suck? What is the general topic? – abatkai May 25 '12 at 18:35
What is inequation? – timur Jun 1 '12 at 6:38

Let $A = \begin{bmatrix} 1-x^2 & xy \\ xy & 1-y^2\end{bmatrix}$. Then $(1-x^2)\xi_1^2+2xy\xi_1\xi_2+(1-y^2)\xi_2^2 = \langle \xi, A\xi \rangle$.

Choose $x = y = \frac{3}{4}$. Then $A = \frac{1}{16}\begin{bmatrix} 7 & 9 \\ 9 & 7\end{bmatrix}$, and choose $\xi = (1,-1)^T$. Then $\langle \xi,A \xi \rangle = -\frac{1}{4}$. So the answer to part of your question is no.

The other side is easier, since $\langle \xi,A \xi \rangle \leq (1-x^2)\xi_1^2+2xy(\xi_1^2+\xi_2^2)+(1-y^2)\xi_2^2 \leq (2-(x-y)^2) (\xi_1^2+\xi_2^2) \leq 2(\xi_1^2+\xi_2^2)$.

Answer to the modified question:

Again choose $\xi = (1,-1)^T$. Now choose $y = x$, and note that we can choose any $|x|<\frac{1}{\sqrt{2}}$. Then we have $\langle \xi,A \xi \rangle = 2(1-2 x^2)$. However, since $\inf_{|x|<\frac{1}{\sqrt{2}}} 2(1-2 x^2) = 0$, no $\lambda>0$ can exist satifying the lower bound $\langle \xi,A \xi \rangle \geq \lambda ||\xi||^2$.

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thanks but it has to be $x^2+y^2<1$ ... I forgot :-( – peter01 May 25 '12 at 19:11
Moving target... – copper.hat May 25 '12 at 19:37

Let $(x,y)\equiv (\cos\phi_0,\sin\phi_0) \;\; $ and $\; (\xi_1,\xi_2)\equiv \rho (\cos\phi,\sin\phi) \;\;$. Then the condition simplifies a lot $(\lambda < \cos^2(\phi-\phi_0)<\mu \;\;)$. It hints that $\lambda=0 \;$ and $\mu=1$. Cheers!

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One asks that $\lambda\|\xi\|^2\leqslant\|\xi\|^2-\langle z,\xi\rangle^2\leqslant\mu\|\xi\|^2$ for every $\xi$, where $\|\ \|$ and $\langle\ ,\ \rangle$ are the usual Euclidean norm and the Euclidean scalar product, and $z=(x,y)$.

Since $\langle z,\xi\rangle^2\leqslant\|\xi\|^2\cdot\|z\|^2$ by Cauchy-Schwarz inequality, the first inequality holds with $\lambda=1-\|z\|^2=1-x^2-y^2$ hence $0\lt\lambda\leqslant1$. The second inequality holds with $\mu=1$.

One can check that these values are optimal, $\lambda$ when $\xi$ and $z$ are colinear and $\mu$ when $\xi$ and $z$ are orthogonal.

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