Find the Fenchel conjugate $f^{*}$ of the function $f:\mathbb{R}\rightarrow ]-\infty,+\infty[$ given by
\begin{equation*}
f(x) =
\begin{cases}
+\infty & \quad \text{if } x\leq0\\
-\ln(x) & \quad \text{if }x>0.\\
\end{cases}
\end{equation*}
My question has to do with the case where $x>0$. I claim the Fenchel conjugate is $-1$ and I am looking for confirmation my result is correct.
proof
Given that $f^*=\sup\langle x,c\rangle-f(x)$ we have
\begin{align*}
f^*&=\sup\langle x,v\rangle+\ln(x)\\
&=\sup\; xv +\ln(x)
\end{align*}
We now have a minimization problem as follows
\begin{align*}
\dfrac{d}{dx}[xv +\ln(x)]=0&\Rightarrow v+\dfrac{1}{x}=0\\
&\Rightarrow x=-\dfrac{1}{v}
\end{align*}
Bakcsubbing this value of x into $xv +\ln(x)$ gives
$$-1+\ln\left(\frac{-1}{v}\right).$$
However $\ln\left(\frac{-1}{v}\right)$ is not defined. Does this mean I can simply disregard it and conclude $f^*(v)=-1\;\forall x>0?$
Thank you for your input.
Best Answer
It depends on whether $v\ge0$ or $v<0$.
First, note $$f^*(v)=\sup_x \langle x,v\rangle -f(x)=\sup_{x>0} \langle x,v\rangle +\ln(x)$$ as $\langle x,v\rangle -f(x)$ equals $-\infty$ when $x\leq 0$.
Second, indeed, when $v\ge 0$, $\ln(-1/v)$ is not defined. But in that case, note, for $x>0$, $$ \frac{d}{dx}[\langle x,v\rangle+\ln x]=v+\frac{1}{x}>0 $$ so $\langle x,v\rangle+\ln x$ is strictly increasing as a function of $x$, and so it achieves its maximum at $x=+\infty$. Then, we deduce $f^*(v)=+\infty$
If $v<0$, you basically proved $f^*(v)=\sup_{x>0} \langle x,v\rangle+\ln(x)=-1+\ln \left(\frac{-1}{v}\right)$