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 have tried to start with this:
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