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Jul 25, 2016 at 3:03 vote accept Oai Thanh Đào
Jul 25, 2016 at 2:15 comment added Iosif Pinelis Yes, this will also do.
Jul 25, 2016 at 2:13 comment added Nate Eldredge Or to skip the convolution, take $n=3$, $m=x_1=0$, and $M=x_2=x_3=1$, and $f(x)=x^k$. Then the inequality reads $$\frac{2}{3} - \left(\frac{2}{3}\right)^k \le \frac{1}{2} - \left(\frac{1}{2}\right)^k$$ which is clearly false for large $k$. (Indeed, for $k \ge 4$.)
Jul 25, 2016 at 2:12 history edited Iosif Pinelis CC BY-SA 3.0
added 401 characters in body
Jul 25, 2016 at 1:59 history answered Iosif Pinelis CC BY-SA 3.0