Suppose I have following function $f(x) = \pi_{i=1}^nx_i$, that is the function is product of all $x_i$. For example, in 2d case, $f(x) = x_1x_2$. I am just wondering whether anyone can help me find the conjugate function of it.
calculating derivating of $y^tx-x_1x_2$ w.r. to $x_1$ and setting it zero gives me $x_2=y_1$ and similarly $x_1 = y_2$, and so my answer is $f^*=y_1y_2$, but don't know if I did it right.
Assuming your domain is $\mathbb{R}^n$.
Let's consider the case where $n=1$.
Let's consider optimization of $xy-x=x(y-1)$ where $y$ is fixed and $x$ is the variable. If $y \ne 1$, we let $x=k(y-1)$ and we can see that we can let $k$ be arbitrarily large.
Hence $$f^*(y)= \begin{cases} \infty &, y \ne 1 \\ 0 &, y=1\end{cases}$$
Now, suppose $n > 1$, $$f^*(y) = \sup_x \{\sum x_iy_i - \prod_{i=1} x_i \}$$
Suppose $y \ne 0$, in particular, we pick an index $j$, such that $y_j \ne 0$,
Let $$x_i = \begin{cases} 0 &, i\ne j \\ ky_j &, i=j\end{cases}$$
Then again, we can let $k$ be arbitarily large.
Also, if $y=0$, we can let $x=(-k, 1, \ldots, 1)$, and $k$ can be arbitrarily large.
$$f^*(y)=\infty$$
