Prove $| u(x) - u_h(x) | \leq h \ \underset{0 \leq y \leq 1}{\max} |u''(y)|$

Prove:

$$| u(x) - u_h(x) | \leq h \ \underset{0 \leq y \leq 1}{\max} |u''(y)|$$

for $$0 \leq x \leq 1$$

Using the fact that

$$|| (u-u_h)' || \leq \underset{0 \leq y \leq 1}{\max} |u''(y)|$$ and the boundary condition $$u(0) = u_h(0) = 0$$

the exercise also suggests

use the relation

$$(u-u_h)(x) = \int_0^x (u-u_h)'(y) dy$$ and the cauchy inequality

where

$$|| w || = (w,w)^{\frac{1}{2}} = ( \int_0^1 w^2 )^{\frac{1}{2}} dx$$

I raise both sides squarely

$$|| (u-u_h)' || = ( \int_0^1 (u-u_h)'^2 dx )^{\frac{1}{2}}$$

getting

$$|| (u-u_h)' ||^2 = | \int_0^1 (u-u_h)'^2 dx |$$

and then reached a dead end

• What is $u_h(x)$? – Robert Z Feb 2 at 18:12
• $u$ is the analytical solution to a pde problem, and $u_h(x)$ is the solution of a finite element problem associated with the same pde – Saiten Feb 2 at 18:35
• I imagine that knowing the exact PDE as well as the equation satisfied by the approximation would be helpful when searching for any relation between them. – Carl Christian Feb 2 at 19:02
• I guess not, beacause $u$ is a general solution as $u_h$ and i guess that is not the way to go. For me its just algebraic work, that i cannot do it btw. But i will give a look from that perspective as well – Saiten Feb 2 at 19:11
• I guess $u_h$ comes from using linear finite elements, correct? – VorKir Feb 3 at 20:34