Linear Functional over $mathbb{R}$ Vector Space

Let us assume to the contrary that there exists $xinker(f)$ such that $xnotinker(g)$. Now $f(x)=0$, so we must have $g(x)ge0$, or more precisely $g(x)>0$ as we have assumed $g(x)ne0$. But then $g(-x)=-g(x)<0$. Although $f(-x)=0$. This contradicts our condition. So $ker(f)subseteqker(g)$.

To prove the reverse inclusion, let us assume there exists $xinker(g)$ such that $xnotinker(f)$. Now take any $y$ such that $g(y)<0$. Such an $y$ must exist, because for any $z$ with $g(z)>0$, $g(-z)<0$, and we have assumed that $g$ is not the zero function. Now let $k>|f(y)|/|f(x)|$. Then if $f(x)>0$, then $f(y+kx)>0$, otherwise if $f(x)<0$, then $f(y-kx)>0$. But $g(y+kx)=g(y-kx)=g(y)<0$. Hence contradicts our condition. So $ker(g)subseteqker(f)$.

Take any $xnotinker(g)$. Let $g(x)=r$. Let $y=r^{-1}x$. Then $g(y)=1$. Also let $alpha=f(y)$. Now as $f$ and $g$ is linear non-zero functionals, so dimension of their image is $1$. By rank nullity theorem we have $V=ker(g)+mathrm{span}(y)$. So any $vin V$ can be written as $v=z+ky$, where $zinker(g)$ and $kinmathbb{R}$. Then $$f(v)=f(z+ky)=f(z)+kf(y)stackrel{(1)}{=}kalpha=alpha(g(z)+kg(y))=alpha g(v)$$ where $(1)$ follows from the fact that $zinker(g)=ker(f)$. Hence $f=alpha g$.