Q:

The dual function of composite functions

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!

A:

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*}

