For fixed nonzero $u,v \in \mathbb{R}^n$ with $u \notin \text{span}(v)$, suppose we have a convex set $E = \{ (x, \alpha) \colon x \in \mathbb{R}^n, x = u + tv, \alpha \in \mathbb{R} \text{ and } R(x, \alpha) \}$ with $R \subset \mathbb{R}^n \times \mathbb{R}$ being some relation. Define $F = \{(t, \alpha) \colon (u+tv, \alpha) \in E\}$.

I know convexity of $F$ can be verified in a straightforward manner. Let us say we want to try another way. My argument for the fact is: define a map $T \colon E \to F$ which is given by $(x, \alpha) \mapsto (t,\alpha)$. If the map is affine, then $F=T(E)$ is clearly convex. For $(x_1, \alpha_1), (x_2, \alpha_2) \in E$, we clearly have $T( (x_1, \alpha_1) + (x_2, \alpha_2) ) = T((x_1, \alpha_1)) + T((x_2, \alpha_2))$.

My question is whether this is sufficient to show $T$ is affine since we know $\beta(x_1, \alpha_1) \notin E$ for $\beta \neq 1$ by our assumption $u \notin \text{span}(v)$. Or is there some other trick allowing us to declear the convexity of $F$? Thanks in advance.

  • $\begingroup$ Isn't $F=\Bbb{R}^2$? Because $t$ and $\alpha$ are independent and can be any real. $\endgroup$ – A.Γ. Nov 28 '17 at 9:11
  • $\begingroup$ @A.Γ.: But $(u+tv, \alpha)$ has to be in $ E$. $\endgroup$ – user1101010 Nov 28 '17 at 17:42
  • $\begingroup$ Maybe I don't understand something, but the set $E$ is all possible pairs $(x,\alpha)=(u+tv,\alpha)$, that is for any $x$ there is some $t$ and for any $t$ there is some $x$. $\endgroup$ – A.Γ. Nov 28 '17 at 18:52
  • $\begingroup$ @A.Γ. You are completely right. Sorry I made some mistake when I abstracted my formation. See my edited version. Thank you. $\endgroup$ – user1101010 Nov 28 '17 at 19:16

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