0
$\begingroup$

I am trying to prove/disprove the following statement:

Let $x_1 \geq \cdots \geq x_n $, $ y_1 \geq \cdots \geq y_n $ be real numbers satisfying $ x_1 + \cdots + x_k \leq y_1 + \cdots + y_k $ for all $ 1 \leq k \leq n$. Let $ f:\mathbb{R} \to \mathbb{R} $ be an increasing convex function. Then one has $ f(x_1) + \cdots + f(x_n) \leq f(y_1) + \cdots + f(y_n) $.

The case $n=1$ is trivial, and the case $n=2$ can be proved as following:

Assume $ x_2 > y_2 $, since otherwise the conclusion follows trivially. Then we can take $ c_1 \geq c_2 \geq 0 $ s.t. $ \begin{cases} f(y_1)-f(x_1) \geq c_1 (y_1-x_1)\\ f(x_2)-f(y_2) \leq c_2(x_2-y_2) \end{cases}$

(for instance, if $f$ is differentiable, then we can take $c_1 = f'(x_1) \geq f'(x_2) = c_2 $.)

Then we see that $f(y_1)+ f(y_2)-f(x_1)-f(x_2) \geq c_1(y_1-x_1) + c_2(y_2-x_2) \geq c_2(y_1+y_2-x_1-x_2) \geq 0 $.

But I have difficulty extending this proof to the case $n\geq 3$. Any help/comment will be appreciated. Thanks.

$\endgroup$
1
$\begingroup$

Let $x_1+x_2+...+x_n=y_1+y_2+...+y_{n-1}+y_n'.$

Thus, $y_n'\leq y_n$, $(y_1,y_2,...,y_n')\succ(x_1,x_2,...,x_n)$ and by Karamata we obtain: $$f(x_1)+f(x_2)+...+f(x_n)\leq f(y_1)+f(y_2)+...+f(y_n')\leq f(y_1)+f(y_2)+...+f(y_n).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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