Conjugate function of negative logarithm I'm trying to understand how conjugate function is built for example like 
$f(x) = -\log x$ where $\operatorname{dom} f = \mathrm{R_{++}}$
One of the lectures which I'm checking
By definition, $f^{*}(y) = \sup_x(yx + \log x)$
Questions:


*

*what does it mean that the difference is bounded?

*why the difference is bounded if and only if $y < 0$?

*why is it reaching its maximum at $x = \frac{-1}{y}$?

 A: You want to compute $\sup_x f(x)$ with $f : \mathbb{R}_{++} \to \mathbb{R}$ defined by $f(x) = yx+\log(x)$. For computing the supremum, it is useful to know the derivative:$$\frac{df}{dx} = y+\frac{1}{x}.$$ Since $x>0$, $df/dx>0$ when $y\geq 0$. That means that when $y\geq 0$:$$\sup_x f(x) = \lim_{x \to \infty} f(x) = \infty.$$
On the other hand, when $y<0$, the derivative  is positive for $x<-1/y$, and negative for $x>-1/y$. That means that the supremum is attained at $x=-1/y$.
A: "The difference" refers to $yx - (-\log x)$, which simplifies to $yx + \log x$. 
 This was because the original function was $f(x) = -\log x$ and you're looking at $yx - f(x)$.
It is bounded iff $y < 0$. You can just do the math to figure this out: you want $\sup_x (yx + \log x)$, so differentiate the inside with respect to $x$ and set it to zero, and you get $y + 1/x = 0$, or $y = -1/x$ as claimed. For $y \geq 0$ this cannot work, since then $x \leq 0$, but that would make $\log x$ undefined.
A: You have to understand that when it says $log(x)$ it really means $ln(x)$
Here's my work below
$$f(x)=-ln(x)$$
$$f'(x)=-1/x$$
Because we know that $f$ is differentiable and this point of maximum gap on $f(x)$ occurs at $f'(x)=y$, therefore the maximum is at 
$-1/x=y$ or $x=-1/y$
So now, to find the $f^*(y)$, we know our slope is $y$, and our x-point is $-1/y$ which makes our y-point $f(-1/y)=-ln(-1/y)$ which also equals $ln(-y)$
Now if I put this in point-slope formula
$$y-(ln(-y))=y(x-(-1/y))$$
Then with $x=0$
$$y-(ln(-y))=y(0-(-1/y))$$
$$y-(ln(-y))=1$$
$$y=ln(-y)+1$$
And therefore my $f^*(y)$ will equal the negative of my $y$ result and I get $-ln(-y)-1$
