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$\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.

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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.

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