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Edit: Consider this solved. I used the wrong definition of linear function. I originally used the polynomial definition, $f(x) = a\cdot x + b$, not the linear map definition $f(\alpha x) = \alpha f(x), f(x + y) = f(x) + f(y)$.

This claim comes from "Convex Optimization" by Boyd and Vandenberghe of recharacterizing a convex optimization problem that has only equality constraints (Section 4.2).

Claim: If a linear function is nonnegative on a subspace then it must be zero on the subspace.

Context: It seems as if this is a general statement, but I've had surprising difficulty trying to prove it in general. The specific problem used is:

Minimize $f_0(x)$ subject to $Ax = b$ where $x \in \mathbb{R}^n$ and $A$ is some $n \times n$ matrix

This claim is applied toward the function $\nabla f_0(x)^T v = \nabla f_0(x) \cdot v \geq 0$ which is a nonnegative linear function over $v \in Nullspace(A) = N(A)$

Work so far: Let $f(x) = a \cdot x + b$ be some nonnegative linear function over some subspace $S$. $0 \in S \rightarrow b \geq 0$. Suppose there exists $x_0 \in S$ such that $a \cdot x_0 + b > 0$. By definition of the subspace, $-x_0 \in S$ such that $-a \cdot x + b \geq 0 \rightarrow b \geq a \cdot x_0 \rightarrow 2b \geq a \cdot x_0 + b > 0$.

By the last relation, if $b = 0$, then contradiction occurs so $b > 0$. At this point, I get the gut feeling I'm doing something wrong.

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    You’re using the wrong definition of *linear function*: in this context $f$ is linear provided that $f(\alpha v)=\alpha f(v)$ for each scalar $\alpha$ and vector $v$. Thus, if $f(v)$ were positive for some $v$ in the subspace, $f(-v)=-f(v)$ would be negative.2017-01-09
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    Kill me. I facepalmed so hard. Thank you for pointing that out. The claim works out without much work. I'm used to referring that version as a "linear map" over linear function.2017-01-09

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Say $f(x)$ is a linear function on a subspace $\mathbb{S}$, then $$ \forall x\in \mathbb{S}, f(x)\geq 0, $$ And $$ -x\in \mathbb{S}, f(-x)\geq 0 \Longrightarrow f(-x)=-f(x)\geq0,\Longrightarrow f(x)\leq 0 $$

Hence, $f(x)=0$.