Let $f(x,\theta)$ be a linear function of $\theta$, meaning $f(x,\theta) = \theta^\top x$, where $\theta$ is a vector or matrix and $x$ is a vector. Is it the case that
$$ \|\theta_1\| \leq \|\theta_2\| \implies \|f(x, \theta_1)\| \leq \|f(x, \theta_2)\|\;? $$
I thought I could prove this using the Cauchy-Schwartz theorem $\|\theta^\top x\| \leq \|\theta^\top\|\|x\|$ but got stuck. Assuming $\|\theta_1\| \leq \|\theta_2\|$ I get $\|\theta_1^\top x\| \leq \|\theta_1^\top\|\|x\|$ and $\|\theta_2^\top x\| \leq \|\theta_2^\top\| \|x\|$. From which I get $\|\theta_1^\top x\|\leq \|\theta_2^\top\|\|x\|$. How do I then get $\|\theta_1^\top x\| \leq \|\theta_2^\top x\|$?
Hints are appreciated.
The answer is "no". A counter example is if $ x $ is in the null space of $ \theta_2 $ but not of $ \theta_1 $.