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The polar of a convex body $C$ is defined as:

For a convex body $C$, its polar $C^*$ is given by $$C^* = \{x\in\mathbb{R}^n: \langle x,c\rangle \le 1, \forall c\in C\}$$

If I start with an orthonormal basis $(e_i)_1^n$, positive constants $(\alpha_j)_1^n > 0$ and the ellipsoid $$\varepsilon = \left\{x\in\mathbb{R}^n: \sum\limits_{j=1}^n \frac{\langle x,e_j\rangle^2}{\alpha_j^2} \leq 1 \right\}$$

how do I show that its polar ellipsoid is $$\varepsilon^* = S_1 = \left\{y\in\mathbb{R}^n: \sum_{j=1}^n \alpha_j^2 \langle y,e_j\rangle^2 \le 1\right\}$$

Using the definition, I saw that $$\varepsilon^* = S_2 = \{z\in\mathbb{R}^n: \langle z,x\rangle \le 1, \text{for all }x\in \varepsilon\}$$ Also, $\langle z,x\rangle = \sum_i\langle z,e_i\rangle\langle x,e_i\rangle$ I used Cauchy Schwartz to do one side of the proof, i.e. $S_1\subseteq S_2$: $$\left(\sum_i\langle y,e_i\rangle\langle x,e_i\rangle\right)^2 \leq \left(\sum_i\alpha_i^2 \langle y,e_i\rangle^2\right)\left(\sum_i \frac{\langle x,e_i\rangle^2}{\alpha_i^2}\right) \le 1 \implies \langle y,x\rangle \le 1$$

I need help with the other direction, i.e. $S_2\subseteq S_1$. Which other inequality might help?

Rodrigo de Azevedo
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stoic-santiago
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1 Answers1

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This is a great deal easier to analyze in the case where you take $C=B_2^n$, the closed Euclidean unit ball. The basic idea is that you take the largest vector contained in the body along the same direction as a given point in the polar. The following proof is in the general case.

For the sake of brevity, denote $\langle x,e_i\rangle$ as $x_i$ for any $x$. Let $z\in S_2$. Define $$v=(\alpha_1 z_1,\alpha_2z_2,\ldots,\alpha_nz_n), v^\circ=\frac{v}{\|v\|},\text{ and }z^\circ = (\alpha_1 v^\circ_1,\alpha_2 v^\circ_2,\ldots,\alpha_n v^\circ_n).$$ It is easily checked that $z^\circ\in C$ (because $v^\circ\in B_2^n$). Applying the fact that $z\in S_2$, we get $$\langle z,z^\circ\rangle\leq 1\implies\langle v,v^\circ\rangle\leq 1 \implies \|v\|\leq 1.$$ Writing this out gives $$\sum_{i} (\alpha_i z_i)^2 \leq 1,$$ which is exactly what we want.

Amit Rajaraman
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