Proof that in case of convex function locally optimal point is also globally optimal
From : CS229, Convex Optimization Overview by Zico Holter, October 2008. http://cs229.stanford.edu/section/cs229-cvxopt.pdf
My question is why : $$\theta f(y) +(1- \theta) f(x)<f(x)$$
For $\theta = 0$ : $$f(x)=f(x)$$
Is it a mistake in notes or I am overlooking something?
Since $y$ is a global minimum while $x$ is not in the contradiction setting,
$$R < \left\| x-y\right\|_2$$
$$\theta = \frac{R}{2\left\| x-y\right\|_2}<1$$
Also, note that $\theta >0$.
Remark: I think your concern is when $\theta = 0$ rather than $\theta = 1$. $\theta$ here is a particular value in $(0 , 1)$.