2
$\begingroup$

I am wondering how FunctionConvexity was implemented in Wolfram Mathematica.

I am trying to prove the convexity of the function $A(v)$:

$$ A(v) = -\dfrac{k-1}{\displaystyle\sum_{i=1}^k \frac{1}{v_i}}, $$

where $k\geq 2$ is some integer constant and $1 \leq v_i \leq k-1$.

The function is easily passed in Mathematica as convex,meaning that it somehow passes the following condition:

$$ f(t x +(1-t)y)\leq t f(x)+(1-t)f(y) ,$$

where $0\leq t\leq 1.$

However, I am having trouble verifying it by hand.

A similar question is also up on math.stackexchange.

$\endgroup$

1 Answer 1

3
$\begingroup$

Here is my guess (Many years of teaching stand behind.). Indeed,

FunctionConvexity[{-(3 - 1)/Sum[1/v[j], {j, 1, 3}], 
v[1] >= 1 && v[2] >= 1 && v[3] >= 1&& v[1] <= 2 && v[2] <= 2 && v[3] <= 2}, {v[1], v[2], v[3]}]

results in 1 which means the function is convex on the cube. I think the command does not check the definition of convexity here, but uses the nonnegativity of its Laplacian by

Minimize[{Laplacian[-(3 - 1)/Sum[1/v[j], {j, 1, 3}], {v[1], v[2], 
v[3]}], {v[1], v[2], v[3]} >= 1&& {v[1], v[2], v[3]} <=2}, {v[1], v[2], v[3]}]

{4/9, {v[1] -> 2, v[2] -> 2, v[3] -> 2}}

The same in other dimensions.

Addition. I think the above is true, however the general approach is more complicated (see Wiki for info):

m = ResourceFunction["HessianMatrix"][-(3 - 1)/
Sum[1/v[j], {j, 1, 3}], {v[1], v[2], v[3]}];
Minimize[{{a, b, c} . m.{a, b, c}, {v[1], v[2], v[3]} >= 1 && {v[1], v[2], v[3]} <= 
2}, {a, b, c}]

{Piecewise[{{0, Inequality[1, LessEqual, v[2], LessEqual, 2] && Inequality[1, LessEqual, v[1], LessEqual, 2] && Inequality[1, LessEqual, v[3], LessEqual, 2]}}, Infinity], {a -> Piecewise[{{0, Inequality[1, LessEqual, v[2], LessEqual, 2] && Inequality[1, LessEqual, v[1], LessEqual, 2] && Inequality[1, LessEqual, v[3], LessEqual, 2]}}, Indeterminate], b -> Piecewise[{{0, Inequality[1, LessEqual, v[2], LessEqual, 2] && Inequality[1, LessEqual, v[1], LessEqual, 2] && Inequality[1, LessEqual, v[3], LessEqual, 2]}}, Indeterminate], c -> Piecewise[{{0, Inequality[1, LessEqual, v[2], LessEqual, 2] && Inequality[1, LessEqual, v[1], LessEqual, 2] && Inequality[1, LessEqual, v[3], LessEqual, 2]}}, Indeterminate]}}

and this means the convexity.

Edit. My guess concerning the Laplacian is not true as $x^2-y^2$ demonstrates. I missed subharmonic and convex functions: explanation, but not justification. The gradient or the Hessian matrix can be used to this end.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer. I am not familiar with the concept of Laplacian, I am reading about it at this very moment. Could be a valid approach for analytical proof? Also, please note that there is a similar question where we are looking for analytical proof on math.stackexchange. $\endgroup$
    – Oleh
    Commented Dec 18, 2021 at 19:24
  • $\begingroup$ I familiarised myself with a concept. However, I guess that the non-negativity of Laplacian implies that the function is element-wise convex. I am not sure that we can conclude that the function is convex over its domain using this method. Please, correct me if I am wrong. $\endgroup$
    – Oleh
    Commented Dec 18, 2021 at 19:55
  • $\begingroup$ @Oleh: You are right. The non-negativity of the Laplacian does not imply the convexity. A simple example is $x^2-y^2$. $\endgroup$
    – user64494
    Commented Dec 18, 2021 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.