On page 134, H. Brezis gives an example of the connection of minimization problem and variational inequality. Here I quote:
Suppose $F$ : $\mathbb R \to \mathbb R$ is a differentiable function and suppose $u \in [0, 1]$ is a point where $F$ achieves its minimum on $[0,1]$. Then either $$u \in (0,1)\text{ and }F′(u) = 0 \tag 1$$ or $$u = 0\text{ and }F′(u) ≤ 0\tag 2$$ or $$u = 1\text{ and }F ′(u) = 1\tag 3$$ These three cases are summarized by saying that $u ∈ [0, 1]$ and $$F ′(u)(v − u) ≤ 0\tag 4$$ $∀v ∈ [0, 1]$
I was really confused by this example. Here are my questions
$(a)$: If I were a high school kid, I would think $(2)$ is wrong. If $F$ achieve the minimium at $0$, should we have $F'(u)\geq 0$?
$(b)$: I couldn't understand how $(3)$ comes from. Why I can have $F'(1)=1$ if $F$ achieve the minimum at $1$?
$(c)$: $(4)$ should be the variational inequality. But to my knowledge, should it be $$ F'(u)(v-u)\geq 0 $$ This fact I use many times in Calculation of variations... Don't tell me it is wrong...
Please help me to understand this example. Thank you!
(for a moment I thought this would be a typo, but I would be surprised if a such good book can make this many consecutive errors...)
You are right on all points.
The corrected version of (4) can be interpreted as saying that the directional derivative in the direction $v - u$ is non negative.