Is the positive function $f(x) = \frac{{{x^D}}}{{x - A}}$ defined in $]A, + \infty[$ with $A \ge 1$ and $D > 1$ a convex function for all these $A$ and $D$ ? I think so but I have a hard time proving it. Thanks.
Asked
Active
Viewed 62 times
1
-
1Have you tried to compute $f''(x)$ ? – Surb Oct 29 '20 at 15:33
-
2What have you tried? – Mark Viola Oct 29 '20 at 15:33
-
Studying the sign of the second derivative becomes prohibitive due to the calculations with $A$ and $D$. The product of two positive functions, convex and with the "same monotony" is a convex function, but $x^D$ is increasing and $\frac{1}{{x - A}}$ is decreaising. – Vincenzo Palmisano Oct 29 '20 at 15:42
-
$D=2$ is a known case – LinAlg Oct 30 '20 at 19:55
1 Answers
0
It is not always convex, as demonstrated by this graph. If it were convex, then the purple dot should have been above the red one.
Zim
- 4,318
-
Thanks, I get it. Now I think that the function is not convex for $ 1 < D <2 $. – Vincenzo Palmisano Oct 31 '20 at 08:28