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. $\endgroup$ – Kavi Rama Murthy 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.)


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