I am given a pde of the following type:
\begin{align} -\Delta f + \vec{a}\cdot \nabla f + f &= g\quad \text{in }\Omega:=B_1(0)\subset\mathbb{R}^2\\ f &= 0\quad \text{on }\partial\Omega \end{align}
with $\vec{a} = (1,1)^t$. For which I have to derive an estimation of the form $||\nabla^2 f||_{L^2} = |\Delta f|_{H^2} \leq c||g||_{L^2}$ (I think it should be the seminorm on $H^2$).
Multiplying by $-\Delta f$ and integrating over the domain yields the following estimate:
\begin{align} ||\Delta f||^2_{L^2} &= - \int g \Delta f dx+ \int f\Delta f dx+ \int \Delta f(\vec{a}\cdot \nabla f)dx\\ &= -(g,\Delta f)_{L^2} +(\vec{a}\cdot\nabla f,\Delta f)_{L^2} - ||\nabla f||_{L^2}^2\\ &\leq |(g,\Delta f)| + |(\vec{a}\cdot\nabla f,\Delta f) - ||\nabla f|||\\ &\leq ||g|| ||\Delta f|| + \frac{1}{2}||\Delta f||^2 \end{align}
here I ommit the $L^2$ subscript in line 3 for brevity. To obtain the last inequality I use the Cauchy-Schwarz inequality on the first term and for the second term the Cauchy inequality:
\begin{equation} (\vec{a}\cdot\nabla f,\Delta f) \leq |(\vec{a}\cdot\nabla f,\Delta f)|\leq ||\vec{a}\cdot\nabla f|| ||\Delta f||\leq \frac{1}{2}||\Delta f||^2 + ||\nabla f||^2 \end{equation}
subtracting $||\nabla f||^2$ on both sides yields the desired estimate.
My first question is: Is this estimation correct?
Now I am given a hint to use the auxiliary problem \begin{align} -\Delta u &= h :=-\Delta f\quad \text{in }\Omega\\ u &= 0 \quad\text{on }\partial\Omega \end{align} to arrive at the desired estimate (which will then be used with the Aubin-Nitsche trick on the auxiliary problem to obtain an estimate of the $L^2$ approximation error).
At this point I am stuck and would like to ask for a hint how to get to the final estimate $||\nabla^2 f||_{L^2} = |\Delta f|_{H^2} \leq c||g||_{L^2}$.
A further guess of mine is:
I know by application of Lax-Milgram, that the original PDE has a solution $f\in H_0^1(\Omega)$. I think I could use a remark from the lecture which says that the Poisson problem (the auxiliary in this case) has a solution in $H^{k+1}$ if $\partial\Omega\in C^k$ and $ h \in H^{k-1}$. A soultion $f\in H^1_0$ does not imply $-\Delta f\in H^{k\geq 0}$ but if I reformulate the Poisson problem in a weak formulation I get $(-\nabla u,\nabla \phi) =(-\Delta f, \phi) \equiv (\nabla f,\nabla\phi)$ and $\nabla f\in H^0=L^2$ thus the remark would be applicable.