Brezis' hint for exercise 8.21.1: how to obtain $1 \leq \|v_0' \|_1$?

Question

I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.21 Assume that $p \in C^1([0,1])$ with $p(x) \geq \alpha>0 \quad \forall x \in[0,1]$ and $q \in$ $C([0,1])$ with $q(x) \geq 0 \quad \forall x \in[0,1]$. Let $v_0 \in C^2([0,1])$ be the unique solution of $$ (1) \quad \left\{\begin{array}{l} -\left(p v_0^{\prime}\right)^{\prime}+q v_0=0 \quad \text { on }[0,1],\\ v_0(0)=1, v_0(1)=0 \end{array} \right. $$

Set $k_0=v_0^{\prime}(0)$.

1. Check that $k_0 \leq-\alpha / p(0)$. [Hint: Multiply equation $(1)$ by $v_0$ and integrate by parts. Use the fact that $1 \leq \|v_0' \|_1 \le \|v_0' \|_2$.]

I could not see how the solution of $(1)$ satisfies $1 \leq \|v_0' \|_1$. Could you please elaborate on this hint?

Answer 1

Notice that $1 = v_0(0) - v_0(1) = - \int_0^1 v_0'$, so $1 = |\int_0^1 v_0'| \le \int_0^1 |v_0'|=\|v_0' \|_1$.