# 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. Csn 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 its zero.

Please correct if I am wrong somewhere.

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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 $f=0$.

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I don't really understand what it means to say that $\lambda$ is arbitrary. But from the above I can deduce that $f = B$. – Shirin Feb 16 '13 at 7:41
Since $f$ is linear, you have $f(\lambda x_0) = \lambda f(x_0)$ for any scalar $\lambda$. From the above you can deduce that $f=0$. Here is another approach: If $f$ is linear, then $f(\mathbb{R}^n)$ is a subspace. However, $\mathbb{R}$ only has two subspaces, itself and $\{0\}$. Since $f$ is bounded below, it must be the latter. – copper.hat Feb 16 '13 at 8:00
Thats awesome.....That does clarify it.....Thanks for a wonderful explanation – Shirin Feb 16 '13 at 8:22
You are very welcome. – copper.hat Feb 16 '13 at 8:23

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 $\lambda\in\mathbb R$?

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$f(\lambda x)$ will be zero only when $\lambda$ is zero. Is that what you meant? If yes I still don't get the point. – Shirin Feb 16 '13 at 2:20
@Shaun See what happens when $|\lambda|$ gets large and $f(x) < 0$. – Dominique Feb 16 '13 at 3:30
I am not sure what will be the implication. I am sorry but I am unable to figure out where you are getting with this....all i can see is that it will cause $\lambda f(x)$ to be some very high positive or negative value..... – Shirin Feb 16 '13 at 3:45

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 $c \in \mathbb{R}$, $$f(c x) = c f(x).$$

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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 $x_0$.

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