Why is supremum of $a*u$ where $\|u\|_2 < r$ is $r \times \|a\|_2$ from the Chebyshev center problem?

Question

Recently I've been reading Convex optimization, in the Chebyshev center problem for linear programming, there is a statement that

$$ \sup\{a^Tu \mid \Arrowvert u\Arrowvert_2 \leq r\} = r\Arrowvert a\Arrowvert_2 \tag 1 $$

where $a$ and $u$ are vectors of $\mathbb R^n$

How to prove this statement?

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I can understand that

$$ \sup\{\|a^Tu\|_2 \mid \Arrowvert u\Arrowvert_2 \leq r\} = r \Arrowvert a\Arrowvert_2 \tag 2$$

But in (1) don't you need the $a > 1$ or $a \succeq I$ ?

Answer 1

Thanks user251257

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By Cauchy–Schwarz inequality,

$$ \|a^Tu\|_2 \leq \|a\|_2 \|u\|_2 = r \|a\|_2 $$

since the right-hand side of (1) is $ r \|a\|_2 \geq 0 $, then (1) holds for both $ a^Tu \geq 0 $ and $ a^Tu \leq 0 $.