Convex conjugate of the exponential, via subdifferentials

Question

Can anyone explain how to go about finding the convex conjugate of $\mathbb{R}\ni x \mapsto e^x$ via the subdifferential convex analysis tricks?

Answer 1

If $f$ is differentiable, then it is true that $f^*(f'(y))= yf'(y)-f(y)$.

Now let $f=\exp$. Then $f'=\exp$ as well and thus $$ f^*(\exp(y))=y\exp(y)-\exp(y).$$ Writing $y=\ln(x) \Leftrightarrow \exp(y)=x$ gives $$f^*(x)=x\ln(x)-x \quad\text{for $x>0$.}$$ Taking $x\to 0^+$ gives $f^*(0)=0\ln(0)-0 = 0$.

It is known from Convex Analysis that the range of $f'$ is dense in the domain of $f^*$. (This result is nontrivial.) Thus the domain of $f^*$ equals $[0,+\infty[$.

We record altogether

$$f^*(x) = \begin{cases} +\infty, &\text{if $x<0$;}\\ 0, &\text{if $x=0$;}\\ x\ln(x)-x, &\text{if $x>0$.} \end{cases}$$