1
$\begingroup$

Consider the real function defined over $x\in (1/4,\infty)$ as follows: $$ f(x)= \begin{cases} x-\frac{7}{4}-\ln (x-3/4) &\qquad\mbox{for }x>3/4\\ x-\frac{5}{4}-\ln (x-1/4) &\qquad\mbox{for }1/4

The function seems to me 'separately convex' for $1/43/4$, and as $x\to (3/4)^-$ approaches a finite limit, while as $x\to (3/4)^+$ diverges. My guess is that I should draw the two tangent lines to $f(x)$ at $x=3/4$ and $x=7/4$ (the argmin of the right branch of the function) and these should ``complete" the convex envelope I am after. The convex envelope should then be composed by $f(x)$ for $1/47/4$, plus the two tangents in between, but I am not sure. Thanks in advance for any clarifying thoughts.

1 Answers 1

1

It seems you have a plot of $f$ at your disposal. For the benefit of the reader I print it here:

enter image description here

The left branch of $f$ ends at the point $P:=\left({3\over4},\log2-{1\over2}\right)$ with slope $-1$ there. Now draw a tangent from $P$ to the right branch. This tangent will have a slope $>-1$ and meets the right branch of $f$ at some point $Q$ near $(1.6,0.05)$. In order to find $Q$ exactly you have to do some analytic geometry; maybe you can solve the resulting equation only numerically.

The convex envelope you are after then consists of the left branch of $f$, the segment $PQ$ and the points of the right branch of $f$ to the right of $Q$.