Showing an inequality on the interpolant error

Question

I have established the inequality $ \|u'-U_{I}'\| \leqslant c_\alpha h\|u''\|$ where $U_{I}$ is a piecewise linear interpolant on $[a,b]$ and $h = \max_{i} x_{i+1} - x_{i}$. Recall the Poincare-Freidrich (PF) inequality:

$$ \|f\| \leqslant c \|f'\|$$ where $f \in \mathcal{H}^{1}(a,b)$ with $f(a) = 0$ or $f(b)=0$. How may I establish the inequality $$ \| u - U_{I}\| \leqslant c_{\beta}h^{2}\|u''\|?$$

I have set $k(x) = u(x) - U_{I}(x)$ to obtain

$$ \| k \| \stackrel{\text{PF}}{\leqslant} c_{1}\|k'\| \stackrel{\text{PF}}{\leqslant} c_1c_2\|k''\| \implies \|k\| \leqslant c\|k''\|$$

but I am having trouble establishing the $h^2$ in this inequality. Thank you.

Answer 1

Usually this is done the same way as you get the first-order for the derivatives. In one-dimensional case you take the interpolant in the explicit form, split the integral in the definition of the norm to a sum over all elements and estimate it. This way you can derive an estimate $$ || u - U_I || \leq c h || u' - U_I' || $$ which in combination with the estimate you already have, will give you the desired result. Edit: Actually you can immediately use the Newton-Leibniz formula, i.e, the fact that $$ (u - U_I) (x) = \int_{x_i}^x (u - U_I)'(y) dy $$ And then bound the $L_2$ norm by the $H^1$ seminorm.