As per the definition of convex function
$f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2) \forall\lambda\in[0,1]\\ \forall x_1,x_2\in\Bbb{R}$
There is a hint for this problem- try to replace $\lambda, x_1,x_2$ with something involving $x,y$ to get that desired inequality. Remember $x<y$
I have also tried using slop condition of convex function, but can't prove it.
Can anyone solve this? Thanks for assistance in advance.

  • 1
    $\begingroup$ I think the inequality should be $f(x) - f(y) \leq f(x-1) - f(y-1)$, otherwise there exists a counterexample. $\endgroup$ – Seewoo Lee Sep 10 at 5:00
  • $\begingroup$ It is still false. $\endgroup$ – Zhaohui Du Sep 10 at 5:02
  • 4
    $\begingroup$ I suspect the inequality is $f(x)+f(y) \le f(x-1) + f(y\color{red}{+}1)$ for $x < y$. If that is the case, it is a special case of Karamata's inequality and above wiki entry has a proof of that. $\endgroup$ – achille hui Sep 10 at 5:04
  • $\begingroup$ @ZhaohuiDu Could you check my answer? $\endgroup$ – Seewoo Lee Sep 10 at 5:07
  • $\begingroup$ @SeewooLee you modification is also true, it is also a special case of karamata inequality. $\endgroup$ – achille hui Sep 10 at 5:12

This is false for $f(x)=x^{2}$.

Hint for the revised question:

$x=\alpha (x-1)+(1-\alpha) (y+1)$ where $\alpha =\frac {y+1-x} {y+2-x}$. Apply the definition of convexity. Similarly we can write $y$ in the form $\beta (x-1)+(1-\beta) (y+1)$. Apply the definition again and add the two inequalities.

  • $\begingroup$ There was a mistake in my question, I have fixed it now. $\endgroup$ – Biswarup Saha Sep 10 at 5:25
  • $\begingroup$ @BiswarupSaha I have edited my answer accordingly. $\endgroup$ – Kabo Murphy Sep 10 at 5:30
  • $\begingroup$ Got it. Nice idea, thank you. $\endgroup$ – Biswarup Saha Sep 10 at 6:45

I'll give a sketch of proof for a modified problem: $f(x) - f(y) \leq f(x-1) - f(y-1)$. This is equivalent to $f(x) - f(x-1) \leq f(y) - f(y-1)$.

Convexity of $f$ is equivalent to the following: for $x_{1} < x_{2} < x_{3}$, $$s(x_{1}, x_{2}) \leq s(x_{1}, x_{3}) \leq s(x_{2}, x_{3})$$ where $$ s(x, y) = \frac{f(y) - f(x)}{y-x} $$ is a slope of the secant line. Using this, we have $$ f(x) - f(x-1) = s(x-1, x) \leq s(x-1, y) \leq s(y-1, y) = f(y) - f(y-1). $$ Equivalence can be proved by replacing $\lambda, x_{1}, x_{2}$ with appropriate variables, which may be in somewhere on Google. (I can't find it now, but I'm sure that it is true.)


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.