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Let $f: \textbf{Y}\rightarrow \bar{\mathbb{R}}$ be a proper closed convex function, $\mathcal{A}: \textbf{E}\rightarrow \textbf{Y}$ be a linear map. Define $g(x)=f(\mathcal{A}x)$. I need to show that $g^*$ is the closed envelope of the function $y\mapsto \inf_x \{f^*(x): \mathcal{A}^*x=y\}$.

Below is what I have: $$f^*(x)=\sup_{\mathcal{A}z} \{\langle x, \mathcal{A}z\rangle-f(\mathcal{A}z)\}=\sup_{\mathcal{A}z} \{\langle \mathcal{A}^*x, z\rangle -g(z)\}=g^*(\mathcal{A}^*x)$$ then $g^*(y)=f^*(x)$ if $\mathcal{A}^*x=y$. However, I'm missing the $\inf_x$ part. Where should I go from here?

kkcocoqq
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  • Isn't $f^*(x)=\sup_{t\in Y}{ \langle x,t\rangle - f(t)}$, rather than $\sup_{y\in \mathcal A[E]} {\langle x,y\rangle-f(y)}=\sup_{z\in E} {\langle x,\mathcal Az\rangle-f(\mathcal A z)}$? –  May 22 '19 at 22:20
  • I thought I could let $t:=Az$ for some $z\in E$. But you're right. $f^(x)$ should be $\geq g^(A^*x)$. – kkcocoqq May 22 '19 at 23:05

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