convex conjugate $X$ 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 we have
\begin{equation}
\sup_{x\in X}\left(\left\langle x^{*},x\right\rangle -\alpha f(x)\right).
\end{equation}
where $a>0$ is a constant. Is it true that in this case we have $\alpha f^{*}(\frac{x^{*}}{\alpha})$?

Yes, since $\sup_x(\langle x^*,x\rangle-\alpha f(x))=\sup_x(\alpha\langle  \frac{x^*}{\alpha},x\rangle-\alpha f(x))=\alpha\sup_x(\langle \frac{x^*}{\alpha},x\rangle- f(x))=\alpha f^*(\frac{x^*}{\alpha})$.
