$X$ is a Banach space and $X^{*}$ denotes its dual. Let $f:X\rightarrow\mathbb{R}$ be an arbitrary convex function. The Fenchel conjugate of $f$ is the function $f^{*}:X^{*}\rightarrow\mathbb{R}$, defined as \begin{equation} f^{*}(x^{*})= \sup_{x\in X}\left(\left\langle x,x^{*}\right\rangle -f(x)\right). \end{equation} My question is how to express the conjugate of f ,when \begin{equation} \sup_{x\in X}\left(\left\langle x,x^{*}\right\rangle -f(\alpha x)\right). \end{equation} where $a\neq 0$ i a given constant.
My idea: We set $\alpha x=y$. Then $x=\frac{y}{\alpha}$. We have
\begin{align*} f^{*}_{\alpha}(x^{*})&= \sup_{x\in X}\left(\left\langle x,x^{*}\right\rangle -f(\alpha x)\right)\\ &=\sup_{y\in X}\left(\left\langle \frac{y}{\alpha},x^{*}\right\rangle -f(y)\right)\\ &=\sup_{y\in X}\left(\frac{1}{\alpha}\left\langle x^{*},y\right\rangle -f(y)\right)\\ &=\sup_{y\in X}\left(\left\langle \frac{x^{*}}{\alpha},y\right\rangle -f(y)\right)\\ &=f^{*}_{\alpha}\left(\frac{x^{*}}{\alpha} \right). \end{align*}
Is the above correct?