The dual function of composite functions

Question

Given $X$ $Y$ are two finite dimensional Hilbert space. Let $K$: $X\to Y$ be linear and $F$: $Y\to \mathbb R^+$ is convex. Let us use $F^\ast$ to denote the dual (conjugate) function of $F$. Recall $$ F^\ast(y^\ast):=\sup\{(y^*,y)-F(y)\} $$ for any $y^\ast\in Y^\ast$. But since Hilbert space, we don't really care $Y^*$ or $Y$...

My question is: what is $((F\circ K)(x))^\ast$ in term of $F^\ast$ and $K^\ast$? Is there a such result? Or just simply $((F\circ K)(x))^\ast = F^\ast (y)$?

Thank you!

Answer 1

Let us assume that $K$ is bounded and invertible (hence, boundedly invertible).

Then, you have \begin{align*} (F \circ K)^* (x^*) &= \sup_x \{ (x^*, x) - (F \circ K)(x) \} \\ &= \sup_y \{ (x^*, K^{-1} y) - F(y) \} \\ &= \sup_y \{ (K^{-*} x^*, y) - F(y) \} \\ &= F^*(K^{-*}(x^*)) = (F^* \circ K^{-*})(x^*). \end{align*}