# A priori FEM estimates without $H^2$ regularity

In basic lectures on finite elements theory, people always assume $$H^2$$ regularity of the solution in order to derive $$O(h^k)$$ a priori estimates in the norms $$H^{2-k}$$, $$k=1,2$$. For simplicity let's restrict to the usual Poisson equation.

I would be interested in a reference of what rates can be expected, in both norms, when the solution is just $$H^1$$ regular (or $$H^s$$ regular), and I am especially curious about approximation properties for the Ritz projection in this less smooth case.

Do you have any hints? Thanks!

Cross posted also on MSE

The usual argument for error estimates in the energy norm is to first use the best-approximation property to get things back to the interpolation error. That is, $$\| u-u_h \|_{H^1} \le C \| u-u_I \|_{H^1}$$ where $$u_I$$ is the interpolant of the solution. For the Laplace equation, $$C=1$$, but for other second-order elliptic equations, $$C$$ might be larger than one.
In any case, then you need to know what happens to the interpolation error on the right in cases of "missing regularity". That is, you are looking for generalizations of the Bramble-Hilbert lemma if you don't have $$H^2$$ regularity. Generally, the literature will tell you that for linear elements, $$\| u-u_I \|_{H^1} \le C_I h^\alpha \| u \|_{H^{1+\alpha}}$$ for all $$0\le \alpha \le 1$$ for which the right hand side makes sense. In other words, if you have $$H^2$$ regularity, you get $$\| u-u_h \|_{H^1} \le CC_I h \| u \|_{H^2},$$ but if you only have $$H^{1+\varepsilon}$$ regularity (e.g., a reentrant corner in your domain with an internal opening angle of strictly less than $$2\pi$$, but not a slit -- for the latter, you would have $$\varepsilon=0$$), then you get $$\| u-u_h \|_{H^1} \le CC_I h^\varepsilon \| u \|_{H^{1+\varepsilon}}.$$ That is, the solution may converge quite slowly if you have a solution that is not very regular.
Since you also ask for convergence in the $$L^2$$ norm: Starting from the estimate above, you then apply the Aubin-Nitsche trick to get things back to the $$L^2$$ norm and that gains you one power of $$h$$.