这是一份nottingham诺丁汉大学ECON1050作业代写的成功案例

Suppose that there exist two distinct points $x_{0}$ and $x_{1}$ of $E$ such that
$$
f\left(x_{1}\right)-f\left(x_{0}\right)=\nabla f\left(x_{0}\right)\left(x_{1}-x_{0}\right) .
$$
Then, by hypothesis,
$$
f(x(\lambda))>\lambda f\left(x_{0}\right)+(1-\lambda) f\left(x_{1}\right) \text { for all } \lambda \in(0,1),
$$
where $\boldsymbol{x}(\lambda)=\lambda x_{0}+(1-\lambda) x_{1}$.
From the above two expressions it follows that
$$
f(x(\lambda))-f\left(x_{0}\right)>(1-\lambda) \nabla f\left(x_{0}\right)\left(x_{1}-x_{0}\right) .
$$
On the other hand, in view of the convexity of $E, \boldsymbol{x}(\lambda)$ is in $E$ for all $\lambda \in(0,1)$. Therefore again,
$$
f(x(\lambda))-f\left(x_{0}\right) \leqq \nabla f\left(x_{0}\right)\left(x(\lambda)-x_{0}\right)=(1-\lambda) \nabla f\left(x_{0}\right)\left(x_{1}-x_{0}\right),
$$
contradicting the inequality established earlier.

ECON1050 COURSE NOTES ：

Let $\boldsymbol{x}$ be any feasible vector. Then $g_{i}(\boldsymbol{x}) \geqq 0=g_{i}\left(\boldsymbol{x}{0}\right)$ for any $i \in I{0}$. Hence,
$$
\nabla_{g_{i}}\left(x_{0}\right) \cdot\left(x-x_{0}\right) \geqq 0 \quad \text { for any } i \in I_{0} \text { and any } x \in F \text {. }
$$
Multiplying both sides of $(58)$ by $\left(\boldsymbol{x}-\boldsymbol{x}{0}\right)$, and applying (64) and (65), we obtain $$ \nabla f\left(\boldsymbol{x}{0}\right) \cdot\left(\boldsymbol{x}-\boldsymbol{x}{0}\right)=-\sum{i \in I_{0}} u_{i 0} \cdot \nabla g_{i}\left(\boldsymbol{x}{0}\right) \cdot\left(\boldsymbol{x}-\boldsymbol{x}{0}\right) \leqq 0
$$
Therefore,
$$
\nabla f\left(x_{0}\right) \cdot\left(x-x_{0}\right) \leqq 0 \quad \text { for every } \boldsymbol{x} \in F
$$
4 Nonlinear Programming
59
Since $f(\boldsymbol{x})$ is pseudoconcave at $\boldsymbol{x}{0},(66)$ implies that $$ f\left(\boldsymbol{x}{0}\right) \geqq f(\boldsymbol{x}) \text { for every } \boldsymbol{x} \in F,
$$
which establishes the optimality of $x_{0}$.