Let $$-\epsilon w_{xx}=f_x-w$$ be given in $\Omega=[0,1]$ for $\epsilon >0$. Moreover, $w=0$ on $\partial \Omega$. It holds $f \in H^1(\Omega)$. So we obtain a unique weak solution $w \in H_0^1(\Omega)$. It holds $$w\in H^2(\Omega)$$
as well as the estimation $$||w||_{H^2(\Omega)} \leq C(\epsilon)||f_x||_{L^2(\Omega)}$$
Now i try to find a explicit specification for $C(\epsilon)$.
Since $w=0$ on $\partial \Omega$ we can use the Poincaré-inequality and we have with $\Omega=[0,1]$
$$||w||_{L^2(\Omega)} \leq ||w_x||_{L^2(\Omega)}$$
Now i try to find $w_x$ in sence of distribution so i can replace this in the inequality.
The weak solution $w$ satisfies the equation in the sense of distributions, that is, for all test functions $\varphi \in C_0^\infty(\Omega)$ holds:
$$\int_\Omega (-\epsilon w_{xx}) \varphi \,dx = \int_\Omega (f_x - w) \varphi \,dx$$
with partial integration it follows
\begin{equation} \int_\Omega w_x \varphi_x \, dx = \frac{1}{\epsilon} \int_\Omega f_x \varphi \, dx - \frac{1}{\epsilon} \int_\Omega w \varphi \, dx. \end{equation}
And now i dont know what to do. Does this mean $$w_x=\frac{1}{\epsilon} f_x - \frac{1}{\epsilon} w \varphi?$$
Then i could find the following for $C(\epsilon)$. $$ ||w||_{L^2(\Omega)} \leq \frac{1}{\epsilon - 1} ||f_x||_{L^2(\Omega)} $$
You can set $\phi=w$. This is permitted by density of $C_c^\infty(\Omega)$ in $H^1_0(\Omega)$. Then $$ \epsilon \|w_x\|_{L^2(\Omega)}^2 + \|w\|_{L^2(\Omega)}^2 \le \|f_x\|_{L^2(\Omega)}\|w\|_{L^2(\Omega)}. $$ And the inequalities $$ \|w\|_{L^2(\Omega)} \le \|f_x\|_{L^2(\Omega)} $$ and $$ \sqrt \epsilon \|w_x\|_{L^2(\Omega)} \le \|f_x\|_{L^2(\Omega)} $$ follow immediately. Using the equation, we have $$ \epsilon \|w_{xx}\|_{L^2(\Omega)} \le \|f_x-w\|_{L^2(\Omega)} \le 2 \|f_x\|_{L^2(\Omega)}. $$