# Line connection function in convex function is convex

$$\newcommand\R{\mathbb R} \newcommand\x{\mathbf x} \newcommand\y{\mathbf y}$$Let $$\gamma\colon\R\to\R^n$$ be the curve defined by $$t\mapsto \x+t(\y-\x)$$, which is a straight line through $$\gamma(0)=\x$$ and $$\gamma(1)=\y$$. Further define $$\widetilde V=V\circ \gamma:\R\to\R$$.

Check that $$\widetilde V$$ is convex given that $$V$$ is convex. I think this is suppose to be just using the definition and readjustment of the terms, but I am not able to do it. Any tips are welcome.

## 1 Answer

$$\newcommand\x{\mathbf x}$$ $$\newcommand\y{\mathbf y}$$ $$\newcommand\R{\mathbb R}$$

Suppose $$a,b\in\R$$. The key to find the correct rearrangement is to realize that $$\gamma$$ will send the convex combination $$ta+(1-t)b)$$, ($$0\leq t\leq 1$$) to a convex combination of the points $$\x+a(\y-\x)$$ and $$\x+b(\y-\x)$$, with the same coefficient $$t$$. The convexity of $$V$$ implies that: \begin{aligned} \widetilde{V}(ta+(1-t)b)&=V(\x+(ta+(1-t)b)(\y-\x))\\ &=V(t(\x+a(\y-\x))+(1-t)(\x+b(\y-\x)))\\ &\leq tV(\x+a(\y-\x))+(1-t)V(\x+b(\y-\x))\\ &=tV(\gamma(a))+(1-t)V(\gamma(b))\\ &=t\widetilde{V}(a)+(1-t)\widetilde{V}(b) \end{aligned} Therefore $$\tilde{V}$$ is also convex.