Let $E$ be a locally convex topological vector space over $\mathbb{R}$. The projectivization $PE$ is the quotient of $E\backslash\{0_{E}\}$ with respect to the equivalence relation $e\sim f$ if $e=\lambda f$.

Is $PE$ a Tychonoff (i.e. completely regular Hausdorff) space?

As far as I can tell, the theorems about the quotient uniform spaces do not apply. On the other hand, it is plausible to expect that this is a known fact.

I can show that $PE$ is completely Hausdorff, i.e. any two points can be separated by a real-valued continuous function. Indeed, if $e\not\sim f$, take $\mu,\nu\in E^{*}$ such that $\left<\mu,e\right>=1, \left<\nu,e\right>=0, \left<\mu,f\right>=0, \left<\nu,f\right>=1$, and consider the map $\mu\oplus \nu:E\to \mathbb{R}^2$. By the definition of the quotient, this map induces a map $\varphi: PE\to P\mathbb{R}^2=S^1$. Since the latter is Tychonoff, we can separate the images of the classes of $e$ and $f$ by a continuous function.