Is it true that a function f is semiconcave if and only if $$ \exists K \;\; s.t. f((1-\lambda)x+\lambda y) \geq (1-\lambda) f(x)+\lambda f(y) + \dfrac{K}{2} \lambda(1-\lambda) |x-y|^2 $$ for any $x,y \in \mathbb{R}^n$ and $\lambda \in [0,1]$ ?
I have found a different definition in "partial differential equations" by Evans i.e. $$ f(x) -\dfrac{C}{2}|x|^2 \;\; \text{concave}.$$
I can't figure it out if it fits to the one above or not.
oh... maybe I have got it right now:
$$ \dfrac{C}{2} |(1-\lambda)x+\lambda y|^2- \dfrac{C}{2}(1-\lambda)|x|^2- \dfrac{C}{2}\lambda|y|^2=\dfrac{C}{2} \left[ ((1-\lambda)x+\lambda y, (1-\lambda)x+\lambda y)-(1-\lambda)(x,x)-\lambda (y,y) \right]=\dfrac{C}{2}\left[(1-\lambda)^2(x,x)+ \lambda^2 (y,y)+ 2\lambda(1-\lambda)(x,y)-(1-\lambda)(x,x)-\lambda (y,y)\right]=\dfrac{C}{2}\left[-\lambda(1-\lambda)(x,x)- \lambda(1-\lambda) (y,y)+ 2\lambda(1-\lambda)(x,y)\right]= -\dfrac{C}{2}\lambda (1-\lambda)\left[(x,x)+(y,y)- 2(x,y)\right]= -\dfrac{C}{2}\lambda (1-\lambda)\left[(x-y,x-y)\right] =-\dfrac{C}{2}\lambda (1-\lambda)|x-y|^2 $$ so that K=-C...isn't it?