Density in $H^1[0,1]$.

Question

Why is $\{u \in C^2[0,1] | u'(0)=u'(1)=0 \}$ dense in $H^1[0,1]$?

Answer 1

Here is a hint - since $C_0^\infty(0,1)$ is dense in $L^2[0,1]$, given $f \in H^1[0,1]$ and $\epsilon > 0$ you can select $v \in C_0^\infty(0,1)$ satisfying $$\|f' - v\|_{L^2[0,1]} < \epsilon/2.$$ Now define $$u(x) = f(0) + \int_0^x v'(t) \, dt$$ and show that $$\|f - u\|_{H^1[0,1]} < \epsilon.$$