# Proof that a point is a local minimum

Theorem: $$\text{Let } U \subseteq \mathbb R. f:U \rightarrow \mathbb R \text{ has a local minimum at } x_0 \text{ if there exists } \epsilon > 0 \text{ such that } f(x_0) \leq f(x) \text{ for all } x \in B_{\epsilon}(x_0) \cap U$$

How would one use this theorem to prove that the function,

$$h(x) = \begin{cases} 5-(x-1)^2 &\text{ if } x \in (0,3),\\ |{x-4}| & \text{ if } x \not\in (0,3). \end{cases}$$

achieves its local minimum at $$x=0$$?

We know that if a function is a local minimum, that the function on either side of the point will be greater than the minimum point, so I was thinking that one could prove that if you took $$f(x_0 + \delta) > f(x_0)$$. I don't get how one would prove that $$5 - (x-1)^2 +\delta$$ > $$5-(x-1)^2$$ is true for $$\delta >0$$, it just seems too trivial, and hasn't proven that the point is a minimum. (Similarly for $$|{x-4}|$$).

Thank you.

• you are taking $f(x_0)+\delta$ instead of $f(x_0+\delta)$ Apr 26, 2020 at 16:33
• Your function is not defined at $0$ nor $3$? Apr 26, 2020 at 16:33
• @Bernard I think the notation is the way it is to make the question I was given more confusing.
– user755706
Apr 26, 2020 at 19:22

I assume you mean that $$h:\mathbb{R}\to\mathbb{R}$$ is defined as $$h(x)=5-(x-1)^2$$ if $$x\in(0,3)$$ and $$h(x)=\lvert x-4\rvert$$ if $$x\in\mathbb{R}\setminus(0,3)$$.
Pick for example $$\delta=1$$. We want to show that for all $$x\in B_\delta (0)$$ we have $$h(x)\geq h(0)=4$$. If $$0, then $$h(x)=5-(x-1)^2\geq 5-(0-1)^2 = 4.$$ If on the other hand $$-1< x < 0$$, then $$h(x)=\lvert x-4\rvert\geq \lvert 0-4\rvert = 4.$$ Hence, $$h$$ obtains a local minimum at 0 with value 4.