If $f$ is a closed proper convex function defined on $\mathbb{R}^n$, prove that the function $\varphi$ defined by $\varphi(\lambda)=f((1-\lambda)x+\lambda y)$, where $x \in \text{dom}f, y \in \mathbb{R}^n$ is a convex function as well. I've tried proving it by brute calculation but it didn't work out well. Any ideas?
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But then $;\phi;$ is defined for fixed $;x,y\in\text{Dom},f;??$ – DonAntonio Jun 21 '16 at 08:29
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Yes, x and y are supposed to be $\mathbb{fixed}$. – Darko M. Jun 21 '16 at 08:35
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Let $\alpha \in [0,1]$.
\begin{align} \varphi(\alpha \lambda_1 + (1-\alpha)\lambda_2)&=f((x+(\alpha \lambda_1 + (1-\alpha)\lambda_2)(y-x)))\\ &=f(\alpha (x+\lambda_1(y-x))+(1-\alpha)(x+\lambda_2(y-x))) \\ &\leq \alpha f(x+\lambda_1(y-x))+(1-\alpha)f(x+\lambda_2(y-x))\\ &=\alpha \varphi(\lambda_1)+(1-\alpha)\varphi(\lambda_2) \end{align}
Siong Thye Goh
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Small detail to add: if $a\lambda_1 + (1-a)\lambda_2$ is a convex combination and $\lambda \in [0,1]$ then $(1-(a\lambda_1 + (1-a)\lambda_2))x + (a\lambda_1 + (1-a)\lambda_2)y$ is a convex combination of x and y. – Piotr Benedysiuk Jun 21 '16 at 09:22