Does there exist two linear functionals $f$ and $g$ on $V$ s.t. $\ker(f)\neq \ker(g)?$

Question

Let $V$ be a finite-dimensional vector space over $\mathbb R$ and let $f$ and $g$ be two non-zero linear functional on $V$ such that whenever $f(x)\geq0,$ we also have $g(x)\geq0.$ Which of the following staements are true (It may also happen that none is true)?

$a.$$\ker(f)\subset\ker(g)$

$b.$$\ker(f)=\ker(g)$

$c.$$f=\alpha g$ for some $\alpha>0.$

As $V$ is a finite-dimensional vector space we can consider $V=\mathbb R^n$ for some $n \in \mathbb N.$ Also from rank-nullity theorem $\dim \ker(f)=\dim \ker(g)$. From here we can say that the above mentioned three options are equivalent.

So can we find two linear functionals on $V$ such that $\ker(f)\neq \ker(g)$ satisfying given conditions.

Answer 1

The answer is that you cannot find $f$ and $g$ satisfying the conditions with $\ker f \ne \ker g$.

Proof: It suffices to prove $\ker f \subseteq \ker g$, since as you noted, $\dim\ker f=\dim \ker g$. Now if $f(v)=0$, then $g(v) \ge 0$ by assumption. However $f(-v)=0$ as well, so $g(-v)\ge 0$. Since both $g(v)$ and $-g(v)$ are nonnegative, we have that $g(v)=0$. Hence $\ker f \subseteq \ker g$. Since their dimensions are equal, this gives us that $\ker f=\ker g$.

Edit in response to N.S. although I've demonstrated what the OP was asking for, namely that if $f$ and $g$ satisfy the conditions then $\ker f=\ker g$, I did so by proving (a) and (b). However, (c) immediately follows from this argument as well, since I proved that if $f$ and $g$ satisfy the conditions, then on the basis $e_1,\ldots,e_n,w$ (where $e_1,\ldots,e_n$ are a basis for the kernel and $w$ is any vector not contained in the kernel) $f$ and $g$ take the values $0,\ldots,0,1$ and $0,\ldots,0,g(w)$ respectively. This demonstrates that $g=g(w)f$. Hence (c) follows immediately.

Answer 2

We show that all three claims are true.

Claim 1: $\ker(f) \subseteq \ker(g)$.

Let $x \in V$ be such that $f(x)=0$. Then $f(\alpha x) \geq 0$ for all $\alpha$ and hence $\alpha g(x) =g( \alpha x) \geq 0 \forall \alpha$. This implies that $g(x)=0$.

Claim 2: $\ker(f) =\ker(g)$. Let $x$ be so that $g(x)=0$.

Assume by contradiction that $f(x) \neq 0$. Pick some $y \in V$ such that $g(y)<0$ (which is possible since $g \neq 0$.

Since $f(x) \neq 0$ there exists some $\alpha$ such that $\alpha f(x) < f(y)$.

Therefore, $f(y -\alpha x) \geq 0$ and hence $g(y -\alpha x)= g(y)- \alpha 0= g(y) \geq 0$. But this contradicts $g(y) \neq 0$.

Claim 3 There exists some $\alpha$ such that $f=\alpha g$.

Fix some $y$ so that $g(y) \neq 0$. Then $f(y) \neq 0$ by Claim 2. Then, there exists some $\alpha$ so that $f(y)=\alpha g(y)$.

We show that $f(x)=\alpha g(x)$ for all $x$.

Let $x \in V$ be fixed but arbitrary.

If $f(x)=0$ the claim follows immediately from claim 2.

If $f(x) \neq 0$, by eventually replacing $x$ y $-x$, we can assume without loss of generality that $f(x) >0$. Then, for all $$\beta > \frac{f(y)}{f(x)}$$ we have $$f(\beta x-y) >0 \Rightarrow g(\beta x-y) >0 \Rightarrow \beta > \frac{g(y)}{g(x)}$$

Taking the infimum of $\beta$ (or equivalently making $\beta \to \frac{f(y)}{f(x)}$) we get $$\frac{f(y)}{f(x)} \geq \frac{g(y)}{g(x)}$$

Since $f(y)=\alpha g(y)$ we get $$\frac{\alpha g(y)}{f(x)} \geq \frac{g(y)}{g(x)}$$

Now, repeat the same argument with $y$ replaced by $-y$:

for all $$\beta > \frac{f(-y)}{f(x)}$$ we have $$f(\beta x+y) >0 \Rightarrow g(\beta x+y) >0 \Rightarrow \beta > \frac{g(-y)}{g(x)}$$

Taking the infimum of $\beta$ (or equivalently making $\beta \to \frac{f(-y)}{f(x)}$) we get $$\frac{f(-y)}{f(x)} \geq \frac{g(-y)}{g(x)}$$

Since $f(y)=\alpha g(y)$ we get $$-\frac{\alpha g(y)}{f(x)} \geq -\frac{g(y)}{g(x)}$$

Combining $$\frac{\alpha g(y)}{f(x)} \geq \frac{g(y)}{g(x)}\\ -\frac{\alpha g(y)}{f(x)} \geq -\frac{g(y)}{g(x)}$$

we get $$\frac{\alpha g(y)}{f(x)} =\frac{g(y)}{g(x)}$$ which gives the claim.