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The exercise is the following. Let $f:U\longrightarrow\mathbb{R}$ be semiconcave on the open set $U$, that is there exists a constant $K\geq0$ such that $$ \lambda f(x)+(1-\lambda)f(y)\leq f(\lambda x+(1-\lambda)y)+\frac{1}{2}K\lambda(1-\lambda)|x-y|^2 $$ for all $x, y\in U$ and $\lambda\in[0, 1]$. Prove that the above inequality is equivalent to the concavity of the map $x\mapsto f(x)-\frac{1}{2}K|x|^2$. I cannot prove this equivalence. In particular, by the concavity of $x\mapsto f(x)-\frac{1}{2}K|x|^2$, I obtain the inequality $$ \lambda f(x)+(1-\lambda)f(y)\leq f(\lambda x+(1-\lambda)y)+\frac{K}{2}(\lambda|x|^2+(1-\lambda)|y|^2-|\lambda x+(1-\lambda)y|^2). $$ How can I get the desired result?

Thank You

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Expanding $$\lvert \lambda x + (1 - \lambda y)\rvert^2 = \lambda^2 \lvert x\rvert^2 + 2\lambda(1 - \lambda)\langle x,y\rangle + (1 - \lambda)^2 \lvert y\rvert^2$$ you can write $$\lambda\lvert x\rvert^2 + (1 - \lambda)\lvert y\rvert^2 - \lvert \lambda x + (1 - \lambda) y\rvert^2 = (\lambda - \lambda^2)\lvert x\rvert^2 - 2\lambda(1 - \lambda)\langle x,y\rangle + [(1 -\lambda) - (1 - \lambda)^2]\lvert y\rvert^2$$ or $$\lambda(1 - \lambda)\lvert x\rvert^2 - 2\lambda(1 - \lambda)\langle x,y\rangle + (1 - \lambda)\lambda \lvert y\rvert^2$$ The latter expression is the same as $$\lambda(1 - \lambda)\lvert x - y\rvert^2$$ You can do something similar for the proof of the other direction.

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