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$X$ 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 $\alpha f^{*}(\frac{x^{*}}{\alpha})$?

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

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  • $\begingroup$ thanks for the answer. I went through the same proof but would like to check with someone else. $\endgroup$ – peter5 Jun 3 '16 at 13:56
  • $\begingroup$ It's totally fine, just be confident yourself :) $\endgroup$ – Peter Jun 3 '16 at 14:04

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