Let $I=(a,b)$, $u\in L^2(I)$ and $\psi\in C^{\infty}(I)$ such that $\psi=0$ on $(a,a+\epsilon)$ and $\psi=1$ on $(b-\epsilon , b)$ for sufficient small $\epsilon$. Let $\varphi\in C_C^\infty (I)$ and consider the function $\theta(x)=\int_{a}^{x}\varphi (y)dy - (\int_{a}^{b}\varphi(y)dy)\psi(x)$, $\theta$ belongs to $C^\infty_C(I) $. My questions are: 1. Why is $\theta'=\varphi$ and 2. is there a $C>0$ such that $\|\theta \|_{L^2}\le C\|\varphi \|_{L^2}$?
I know that $\theta'(x)=\varphi(x)-(\int_{a}^{b}\varphi(y)dy)\psi'(x)$ but why should $(\int_{a}^{b}\varphi(y)dy)\psi'(x)$ be zero?
It isn't, the derivative of $\theta$ really has those two terms
No. Actually there isn't even any reason why $\theta$ should be square integrable, given its dependence on $\psi.$ But even if it is, then its quadratic norm is stronly influenced by the choice of $\psi.$ However, if $\psi$ is fixed and square integrable and we only vary $\varphi$ then
\begin{eqnarray} \|\theta\|_2&\leq&\|\varphi\|_1+\|\varphi\|_1\|\psi\|_2 \\ &\leq&\|\varphi\|_2\sqrt{b-a}(1 +\|\psi\|_2)\\ \end{eqnarray}