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Let $v: \mathbb{R} \to \mathbb{R}$ be an increasing, convex function.

For any $t>0$ I want to show that for all $x_{1} \leq x_{2}$ we have: $$v(x_{1}+t) - v(x_{1}) \leq v(x_{2} +t) - v(x_{2})$$

This of course can be illustrated heuristically if $v$ is twice differentiable. But I am trying to show this from the definition of a convex function and by the fact that $v$ is increasing, but I am just moving in circles. I am pretty sure this result is true, and I need it to finish a proof I am working on.

Any suggestions will help.

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Since $x_1 \le x_1+t,x_2 \le x_2+t$, from the definition of convexity we have $$ v(x_1+t) \le \left(\frac{x_2-x_1}{x_2-x_1+t}\right)v(x_1) + \left(\frac{t}{x_2-x_1+t}\right)v(x_2+t) $$ and $$ v(x_2) \le \left(\frac{t}{x_2-x_1+t}\right)v(x_1) + \left(\frac{x_2-x_1}{x_2-x_1+t}\right)v(x_2+t).$$ Adding up the inequalities gives the desired inequality.

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  • $\begingroup$ I think this is a great and succinct answer. However, before I accept and upvote, I am wondering if you can explain the steps you took to get to this solution? I worked for a while on this problem and could not get here, so I want to understand how you reached this point. $\endgroup$ – möbius Oct 9 '15 at 18:33
  • $\begingroup$ Nevermind, I see the logic. Thanks for your help. $\endgroup$ – möbius Oct 9 '15 at 18:42
  • $\begingroup$ I'm sorry, i made mistake for edit. Original answer is also correct. It is good answer. $\endgroup$ – hew May 22 at 14:09
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[This inequality is true for any convex function, increasing or not.]

I'm going to assume that $x_2\le x_1+t$. The reasoning in the remaining case is similar. First consider the three points $x_1\le x_2\le x_1+t$. We have $$ x_2={x_1+t-x_2\over t}x_1+{x_2-x_1\over t}(x_1+t), $$ so by the convexity inequality for $\nu$, $$ \nu(x_2)\le {x_1+t-x_2\over t}\nu(x_1)+{x_2-x_1\over t}\nu(x_1+t). $$ Similarly, $$ \nu(x_1+t)\le {x_2-x_1\over t}\nu(x_2)+{x_1-x_2+t\over t}\nu(x_2+t). $$ Now add these two inequalities, clear $t$ from the denominator, and simplify.

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