I am interested in learning more about optimization problems. I came accross a theorem stating that a local minimum of a convex function is the global minimum of that function. I derived the following proof, but I am not sure whether it is correct and would like to receive some feedback on it.
A function $f:S\rightarrow\mathbb{R}$ for which $S\subset\mathbb{R}^{n}$ is convex if for all $x,y \in S$ it holds that $f(\alpha.x + (1-\alpha).f(y) ) \le \alpha.f(x) + (\alpha - 1).y$ where $\alpha \in \lbrack 0,1\rbrack$.
Assume by contradiction that $x_{0}$ is the local minimum of $f$ but not the global minimum. This implies that there must exist a $x^{*}$ so that $x^{*} < x_{0}$.
Now consider the points described by $\alpha.x_{0} + (1-\alpha).x^{*}$ where $\alpha \in \lbrack 0,1\rbrack$. There must exist at least one $\alpha_{0} \in \lbrack 0,1\rbrack$ so that $f(\alpha_{0}.x_{0} + (1-\alpha_{0}).x^{*} ) > \alpha_{0}.f(x_{0}) + (1-\alpha_{0}).f(x^{*})$. Would this be not the case then $x_{0}$ would not be a local minimum. $f(\alpha_{0}.x_{0} + (1-\alpha_{0}).x^{*} ) > \alpha_{0}.f(x_{0}) + (1-\alpha_{0}).f(x^{*})$ is in violation with the convexity of $f$ and therefore we have proofed the theorem by contradiction.