To make $\max_a c^T a - \alpha\cdot\|a\| < \infty$, what is the requirement for $c$ and $\alpha$

Question

To make $\max_a c^T a - \alpha\cdot\|a\| < \infty$, where $a$ and $c$ is a vector, $\alpha$ is a scalar. $a$ can go infinitely large.

My question is what is the requirement for $c$ and $\alpha$, in order to make $\max_a c^T a - \alpha\cdot\|a\|$ not go to infinte.

I know the answer is $\|c\|\leq \alpha$, but why?

Answer 1

Consider Cauchy-Schwartz inequality $|c^Ta|\leq \|c\|\|a\|$. Hence, we have $$ \max_a c^Ta-\alpha \|a\| \leq \max_a \left(\|c\|-\alpha \right)\|a\|. $$ Since $a$ can go infinitely large, then $\|c\| \leq \alpha$ must be hold, which makes the object function non-positive. On the contrary, if $\|c\| > \alpha$, we set $a=\lambda c$ and let $\lambda$ tends to infinity to derive a contradiction.

Answer 2

I think I have figured it out!

Step 1: Let the two vectors and one scalar in the expression be fixed, i.e., $c, a, \alpha$ is fixed. We have

$$ c^T a - \alpha \|a\| \leq \|c\|\|a\| - \alpha\|a\|,$$

that is because $c^Ta = \|c\|\|a\|\cos\theta \leq \|c\|\|a\|$. Hence, we only need to make $$\max_a \|c\| \|a\| - \alpha \cdot\|a\| = \max_a (\|c\|-\alpha)\|a\|$$

not going to infinity.

Step 2: Since $\|\alpha\|$ can go infinity large, we have to make $\|c\|-\alpha$ negative, that is $$\|c\|-\alpha \leq 0 $$

Finish the result.