Bounded linear function implication In Stephen Boyd's book, Boyd uses the theorem that a linear function is bounded below  on $R^m$ only when it is zero. I can't really digest this. Can someone tell me why this holds?
I mean if I take a line, convert it into a ray. It can start at any point. So indeed its bounded below and its linear but it doesn't mean that it's zero.
Please correct if I am wrong somewhere.
 A: Suppose $f$ is linear, and $f(x) \geq B$ for all $x$. Choose any $x_0$. Then  $f(\lambda x_0) = \lambda f(x_0) \geq B$ for all $\lambda$.
Take $\lambda >0$, which gives $f(x_0) \geq \frac{B}{\lambda}$, and since $\lambda$ is arbitrary, we have $f(x_0) \geq 0$.
Now take $\lambda <0$, which gives $f(x_0) \leq \frac{B}{\lambda}$, which now results in $f(x_0) \leq 0$.
Together, this gives $f(x_0) = 0$. Since $x_0$ was arbitrary, we have $f=0$.
A: Following the above, $f(x_0) \geq \frac{B}{\lambda}$ and $f(x_0) \geq B$. 
For $\lambda > 0$. 
If $f(x_0) < 0$, then $\frac{B}{\lambda}$ could be greater than $f(x_0)$ for some $\lambda$, which is a contradiction. Hence, $f(x_0) \geq 0$. 
For $\lambda \leq 0$. 
If $f(x_0) \leq 0$, then $\lambda f(x_0) \geq 0 \geq B$ holds. Hence, $f(x_0) \leq 0$.
As desired, $f(x_0) = 0$ for any $x_0$.  
A: Hint: If $f$ is a non-zero linear function then there exists a point $x$ such that $f(x)\neq 0$. What can you say about $f(\lambda x)=\lambda f(x)$ for $\lambda\in\mathbb R$? 
A: I think you're getting tripped up on the fact that your linear function $f : \mathbb{R}^m \to \mathbb{R}$ is defined on all of $\mathbb{R}^m$. You cannot consider $f$ on a restricted domain (e.g., picking a ray from a line). Since $f$ is linear, we have the property that for all $x \in \mathbb{R}^m$ and all $c \in \mathbb{R}$, $$f(c x) = c f(x).$$
