这是一份manchester曼切斯特大学ECON20072T作业代写的成功案例

$$
f\left(x_{0}\right)-f\left(\lambda x_{0}+(1-\lambda) x_{1}\right) \leqq(1-\lambda) \cdot \nabla f\left(\lambda x_{0}+(1-\lambda) x_{1}\right) \cdot\left(x_{0}-x_{1}\right)
$$
and
$$
f\left(x_{1}\right)-f\left(\lambda x_{0}+(1-\lambda) x_{1}\right) \leqq-\lambda \cdot \nabla f\left(\lambda x_{0}+(1-\lambda) x_{1}\right) \cdot\left(x_{0}-x_{1}\right) .
$$
Therefore
$$
\lambda f\left(x_{0}\right)+(1-\lambda) f\left(x_{1}\right) \leqq f\left(\lambda x_{0}+(1-\lambda) x_{1}\right) \text { for every } \lambda \in[0,1] .
$$

ECON20072T COURSE NOTES ：

Hence, from the assumed quasi-concavity of $g_{\AA}(x)$,
$$
g_{j}\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \geqq 0 \quad(j=1, \ldots, m) \quad \text { for any } \lambda \in[0,1] .
$$
which shows the convexity of $F$. Since $F$ is convex and $x_{0}$ is optimal for (I), for any $x \in F$ and every $\lambda \in[0,1]$ we have
$$
f\left(x_{0}\right) \geqq f\left(\lambda x+(1-\lambda) x_{0}\right)=f\left(x_{0}+\lambda\left(x-x_{0}\right)\right) .
$$
In view of the differentiability of $f$ and the optimality of $x_{0}$, we have further $\nabla f\left(x_{0}\right) \cdot\left(x-x_{0}\right)+\alpha\left(x_{0}, \lambda\left(x-x_{0}\right)\right) \cdot\left|x-x_{0}\right| \leqq 0 \quad$ for any $\lambda \in(0,1] .$