# Local minima are global for convex functions

Let $$f:\mathbb R^n\to \mathbb R$$ be a convex function.

Show that any local minimum point is also a global minimum point.

I've seen a proof by contradiction here(end of pp.2), but I'd like to prove statements directly whenever I can.

The idea is simple, because $$x_0$$ is a local minimum point, we have a neighborhood $$V$$ of $$x_0$$ satisfying $$\forall x\in V, f(x_0)\leq f(x)$$ Pick $$x\in\mathbb R^n$$ and take the convex combination s.t $$\lambda x_0 + (1-\lambda) x\in V$$, then $$f(x_0) \leq f(\lambda x_0 + (1-\lambda) x) \leq \lambda f(x_0) + (1-\lambda)f(x)\xrightarrow[\lambda\to 0]{}f(x)$$ The limit exists because addition of vectors and multiplication with scalar are continuous operations. But surely, when $$\lambda \to 0$$ we potentially exit the set $$V$$. Are we allowed to conclude $$f(x_0)\leq f(x)$$? If so, why isn't it a problem that the convex combinations along the path $$\lambda \to 0$$ aren't all in $$V$$?

Take any $$x \in \mathbb{R}^n$$ and choose $$\lambda \in (0,1)$$ such that $$y:= \lambda x_0 + (1-\lambda)x \in V$$. Then $$f(y) \leqslant \lambda f(x_0) +(1-\lambda)f(x)$$ by convexity. But $$f(x_0) \leqslant f(y)$$, so we have $$f(y) \leqslant \lambda f(y) + (1-\lambda)f(x)$$, thus $$f(y) \leqslant f(x)$$, so $$f(x_0) \leqslant f(x)$$.