1) Prove that T(K) = [-A,A] where A = $\sqrt{\sum_{i=1}^p (a_i)^2}$

Question

Let T: $\mathbb R^p \to \mathbb R $ such that $ T(X)= a_1x_1 + a_2x_2 + .....a_px_p$ with ${A_v} = (a_1,a_2,...a_n) \in \mathbb R^p. $

Let K = set of ${X \in \mathbb R^p : ||X|| <1} $

1) Prove that T(K) = [-A,A] where A = $\sqrt{\sum_{i=1}^p (a_i)^2}$

For this I started with $\sqrt{\sum_{i=1}^p (a_i)^2(x_i)^2} < \sqrt{\sum_{i=1}^p (a_i)^2} \sqrt{\sum_{i=1}^p(x_i)^2} < \sqrt{\sum_{i=1}^p (a_i)^2} $

Hence $ -\sqrt{\sum_{i=1}^p (a_i)^2} < T(X) < \sqrt{\sum_{i=1}^p (a_i)^2} $

Is this sufficient to prove 1?

2) Obtain $x_*$ and $x^* $ such that $T(x_*)$ = -A and $T(x^*)$ = A

I do not understand how I should start with this part of the problem. Should I set the inverse of T and determine the values?

Answer 1

What you proved in 1), if you use "less than or equal" is that $T(K)\subset[-A,A]$. To show the reverse inclusion, you need to show that $-A\in T(K)$, and $A\in T(K)$, and that $T(K)$ is convex.

For 2), take $$ x^*=\left(\frac{a_1}{\sqrt{\sum_ja_j^2}},\ldots,\frac{a_p}{\sqrt{\sum_ja_j^2}}\right), $$ and $$ x_*=\left(-\frac{a_1}{\sqrt{\sum_ja_j^2}},\ldots,-\frac{a_p}{\sqrt{\sum_ja_j^2}}\right), $$