MHB How can the convexity of a disk be proved using linear algebra?

[rputra]

I am working on a linear algebra problem like this:

> Consider the set of all points $(x,y) \in \mathbb R^2$ as defined by $x^2 + y^2 \leq 25$. Prove that $x^2 + y^2 \leq 25$ is convex.

Here is what I have made out so far:

(1) $x^2 + y^2 \leq 25$ is a disk with center at the origin of Cartesian coordinate with radius of $5$. Consider two points in the disk as defined by $P(a,b)$ and $Q(c,d)$. Consider also point $R(e,f)$ which lays in the segment line $\overline {PQ}$.

(2) By way of contradiction, we will assume that $R(e,f)$ lays outside of the disk. Hence, while we have $a^2+b^2 \leq$ 25 and $c^2+d^2 \leq 25$, we assume that $e^2+f^2 > 25$.

(3) Since $R(e,f)$ lays in the line segment $\overline {PQ}$, there exist $t_1 \geq 0$, and $t_2 \geq 0$, with $t_1 + t_2 = 1$, such that
$$\begin{align}
(e, f) &= t_1(a,b) + t_2(c,d)\\
&= at_1 + ct_2, bt_1 + dt_2.
\end{align}$$

After the third step, I believe that I need to algebraically manipulate $(at_1 + ct_2)^2 + (bt_1 + dt_2)^2$ such that $(at_1 + ct_2)^2 + (bt_1 + dt_2)^2 > 25$ is impossible. However, I am stuck on how to contradict my own assumption. Do I have to use the Triangle Inequality instead? Any helps or suggestion would be very much appreciated.

Thank you before hand for your time and effort.

 

[Euge]

Hi Tarrant,

To show that your set (which I'll call $\Omega$) is convex, prove that given $\lambda\in [0,1]$, $(x,y),(a,b)\in \Omega$, we have $\lambda(x,y) + (1 - \lambda)(a,b)\in \Omega$, i.e., $(\lambda x + (1 - \lambda)a, \lambda y + (1 - \lambda) b) \in \Omega$. Obtain the identity

$$[\lambda x + (1 - \lambda)a]^2 + [\lambda y + (1 - \lambda)b]^2 = \lambda^2(x^2 + y^2) + 2\lambda(1 - \lambda)(xa + yb) + (1 - \lambda)^2(a^2 + b^2)$$

Since $x^2 + y^2 \le 25$ and $a^2 + b^2 \le 25$, then $\lambda^2(x^2 + y^2) \le 25\lambda^2$ and $(1 - \lambda)^2(a^2 + b^2) \le 25(1 - \lambda)^2$. Now, by the Cauchy-Schwarz inequality, $xa + yb \le \sqrt{x^2 + y^2}\cdot \sqrt{a^2 + b^2} \le 5 \cdot 5 = 25$. Therefore $2\lambda(1 - \lambda)(xa + yb) \le 25 \cdot 2\lambda(1-\lambda)$. Consequently,

$$\lambda^2(x^2 + y^2) + 2\lambda(1 - \lambda)(xa + yb) + (1 - \lambda)^2(a^2 + b^2) \le 25\lambda^2 + 25\cdot 2\lambda(1-\lambda) + (1 - \lambda)^2 = 25[\lambda^2 + 2\lambda(1 - \lambda) + (1 - \lambda)^2] = 25[\lambda + (1 - \lambda)]^2 = 25$$

This proves $[\lambda x + (1 - \lambda)a]^2 + [\lambda y + (1 - \lambda)b]^2 \le 25$, as desired.

 

[rputra]

Hi Tarrant,

To show that your set (which I'll call $\Omega$) is convex, prove that given $\lambda\in [0,1]$, $(x,y),(a,b)\in \Omega$, we have $\lambda(x,y) + (1 - \lambda)(a,b)\in \Omega$, i.e., $(\lambda x + (1 - \lambda)a, \lambda y + (1 - \lambda) b) \in \Omega$. Obtain the identity

$$[\lambda x + (1 - \lambda)a]^2 + [\lambda y + (1 - \lambda)b]^2 = \lambda^2(x^2 + y^2) + 2\lambda(1 - \lambda)(xa + yb) + (1 - \lambda)^2(a^2 + b^2)$$

Since $x^2 + y^2 \le 25$ and $a^2 + b^2 \le 25$, then $\lambda^2(x^2 + y^2) \le 25\lambda^2$ and $(1 - \lambda)^2(a^2 + b^2) \le 25(1 - \lambda)^2$. Now, by the Cauchy-Schwarz inequality, $xa + yb \le \sqrt{x^2 + y^2}\cdot \sqrt{a^2 + b^2} \le 5 \cdot 5 = 25$. Therefore $2\lambda(1 - \lambda)(xa + yb) \le 25 \cdot 2\lambda(1-\lambda)$. Consequently,

$$\lambda^2(x^2 + y^2) + 2\lambda(1 - \lambda)(xa + yb) + (1 - \lambda)^2(a^2 + b^2) \le 25\lambda^2 + 25\cdot 2\lambda(1-\lambda) + (1 - \lambda)^2 = 25[\lambda^2 + 2\lambda(1 - \lambda) + (1 - \lambda)^2] = 25[\lambda + (1 - \lambda)]^2 = 25$$

This proves $[\lambda x + (1 - \lambda)a]^2 + [\lambda y + (1 - \lambda)b]^2 \le 25$, as desired.

Thank you for your quick response, and apology for not responding fast enough. I have one question though about the Cauchy-Schwartz Inequality:

(1) According to Wolfram MathWorld, the vector form of the inequality looks like this:
$$||\mathbf a \cdot \mathbf b|| \leq ||\mathbf a|| \ ||\mathbf b||,$$
see its link here: Cauchy's Inequality -- from Wolfram MathWorld

Don't you think the left-hand side should be absolute sign instead, so that it goes like this:
$$|\mathbf a \cdot \mathbf b| \leq ||\mathbf a|| \ ||\mathbf b||,$$
simply because the result of a dot product is already scalar?

(2) If my observation is true, do you think you have to make sure that the left-hand side is always positive in order for your proof to be valid? Please let me know.

Thank you again and again for your quick response.

 

[Euge]

Thank you for your quick response, and apology for not responding fast enough. I have one question though about the Cauchy-Schwartz Inequality:

(1) According to Wolfram MathWorld, the vector form of the inequality looks like this:
$$||\mathbf a \cdot \mathbf b|| \leq ||\mathbf a|| \ ||\mathbf b||,$$
see its link here: Cauchy's Inequality -- from Wolfram MathWorld

Don't you think the left-hand side should be absolute sign instead, so that it goes like this:
$$|\mathbf a \cdot \mathbf b| \leq ||\mathbf a|| \ ||\mathbf b||,$$
simply because the result of a dot product is already scalar?

Yes, you're right. The author made a typo there.

(2) If my observation is true, do you think you have to make sure that the left-hand side is always positive in order for your proof to be valid? Please let me know.

No. For all real numbers $t$, $t\le \lvert t\rvert$. So $xa + yb \le \lvert xa + yb\rvert \le \sqrt{x^2 + y^2}\sqrt{a^2 + b^2}$. Since $2\lambda(1 - \lambda)$ is a nonnegative number, then $2\lambda(1 - \lambda)(xa + yb) \le 2\lambda(1-\lambda)\sqrt{x^2 + y^2}\sqrt{a^2 + b^2}$.

 

[rputra]

Thank you again and again.