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Let $E = \{x \in\mathbb R^p : \sum_{i = p}^p X_i^2/\alpha_i^2 \leq 1\}$ Prove that $E$ is closed and bounded.

To prove that $E$ is closed I used the fact that the boundary of the set $E$ is equal to $\{x \in\mathbb R^p : \sum_{i = p}^p X_i^2/\alpha_i^2 = 1\}$ and the boundary is contained in the $E.$ So $E$ is closed.

However I do not know how to prove the fact that it is bounded.

Michael Hardy
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FSteval
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    Hint: If your notation means what I think it means, then notice that this is the equation of a filled in hyperellipsoid. This lies inside the ball of radius equal to the length of the semi major axis. Let $\alpha = max( \alpha_i )$, then $E \subset E'$ where $E' = {x \in \mathbb{R}^p : \sum_{i=1}^p X_i^2/\alpha^2 \leq 1}$ and so lies within a ball of radius $\alpha$ so is bounded. – Osama Ghani May 19 '18 at 18:15
  • So if I choose the maximum{$\alpha_i$} then E is contained in the sum of the elements taking the maximum? – FSteval May 19 '18 at 18:18
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    Yes! Note the implication that $\sum_{i=1}^p X_i^2/\alpha_i^2 \leq 1 \implies \sum_{i=1}^p X_i^2/\alpha^2 \leq 1$ since increasing the denominator on each term ensures the sum is less than or equal to before. Or you could see this geometrically as the equation of a hyperellipsoid inside a hypersphere. – Osama Ghani May 19 '18 at 18:20
  • @nicomezi No that wouldn't work. Take the case $\frac{x^2}{2^2} + y^2 \leq 1$. This does not lie within the set $x^2 + y^2 \leq 1$. If we follow the logic of the math above, choosing a max $\alpha$ corresponds to dividing by a bigger number so each term is less than or equal to before, so the inequality still holds. – Osama Ghani May 19 '18 at 18:22
  • Note my edits to the question. In the expression $$ {x \in\mathbb R^p : \sum_{i = p}^p X_i^2/\alpha_i^2 = 1} $$ the ${\text{curly braces}}$ should be inside of MathJax, as should the expression $E={}.$ And I changed $R^p$ to $\mathbb R^p.$ Are you taking it to be conventional that the scalar components of a vector denoted by a lower-case $x$ are denoted by capital $X_i \text{ ?} \qquad$ – Michael Hardy May 19 '18 at 18:39
  • @OsamaGhani Thank you. I thought that you couldn't put a ball inside an ellipsoid, but by placing the radius as the maximum semi-axis you guarantee it – FSteval May 19 '18 at 18:57
  • Thank you @MichaelHardy, I will follow such format from now on. Yes, that is the convention I have followed. – FSteval May 19 '18 at 18:59

3 Answers3

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To see that the set

$E = \left \{ x = (X_1, X_2, \ldots X_p) \in \Bbb R^p \mid \displaystyle \sum_1^p \dfrac{X_i^2}{\alpha_i^2} \le 1 \right \} \tag 1$

is bounded, let

$\mu = \max \{ \vert \alpha_i \vert, \; 1 \le i \le p \}; \tag 2$

then

$\dfrac{1}{\mu^2} \le \dfrac{1}{\vert \alpha_i \vert ^2} = \dfrac{1}{\alpha_i^2}, \; 1 \le i \le p, \tag 3$

whence

$\dfrac{X_i^2}{\mu^2} \le \dfrac{X_i^2}{ \vert \alpha_i \vert^2} = \dfrac{X_i^2}{\alpha_i^2}, \; 1 \le i \le p; \tag 4$

then for $x = (X_1, X_2, \ldots, X_p) \in E$ we have

$\displaystyle \sum_1^p \dfrac{X_i^2}{\mu^2} \le \sum_1^p \dfrac{X_i^2}{\alpha^2} \le 1, \tag 5$

and thus, upon multiplication by $\mu^2$,

$\displaystyle \sum_1^p X_i^2 \le \mu^2, \tag 6$

which shows that every $x \in E$ lies in the sphere of radius $\mu$ centered at the origin; thus $E$ is a bounded set.

As for $E$ being closed, the easiest approach here, I think, is to show that $\bar E$, the complement of $E$, is open; to see this, we use the fact that the function

$f(x) = \displaystyle \sum_1^p \dfrac{X_i^2}{\alpha_i^2} \tag 7$

is continuous, which I think is quite evident; then clearly

$E = \{x \in \Bbb R^p \mid f(x) \le 1 \} = f^{-1}([0, 1]), \tag 8$

so

$\bar E = \{x \in \Bbb R^p \mid f(x) > 1 \} = f^{-1}((1, \infty)); \tag 9$

since the inverse image of an open set under a continuous mapping is open, we see that $\bar E$ is open in $\Bbb R^p$, whence $E$ is closed, being the complement of the open $\bar E$.

Robert Lewis
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Let $\alpha = max(\alpha_i)$ then if E is as you previously described it, then let $E_2$ be the ball of radus equal to the length of the major semi-axis ( defined as $max(\alpha_i)$. Clearly E is a subest of $E_2$ so it is then bounded

Ana Pau De la Fuente
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Try to get a lower bound on $\sum\frac{x_i^2}{\alpha_i^2}$ involving the Euclidean norm of $x$. For example:

$$\sum\frac{x_i^2}{\alpha_i^2}\geq\min_i\left(\frac1{\alpha_i^2}\right)||x||_2^2$$

Jack M
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