I want to show that the following Linear operator $L$ is bounded and surjective:
$L: H^2(I) \to \mathbb{C}^4$ where $I = [\alpha,\beta]\subset \mathbb{R}$
given by $L(u) = (u(\alpha),u(\beta),u'(\alpha),u'(\beta))$
I have some trouble with this: The Sobolev space $H^2(I)$ is endowed with the m-norm: see http://en.wikipedia.org/wiki/Sobolev_space
In this case $$\left\|u\right\| = \left( \int_{\alpha}^{\beta} |u|^2dx \right)^{\frac{1}{2}}+ \left(\int_{\alpha}^{\beta}|u'|^2dx\right)^{\frac{1}{2}} + \left(\int_{\alpha}^{\beta}|u''|^2dx\right)^{\frac{1}{2}} $$
So we must show that for some constant $C$ we have $$(|u(\alpha)|^2+|u(\beta)|^2+ |u'(\alpha)|^2+|u'(\beta)|^2)^{\frac{1}{2}} \leq C \left\|u\right\| $$
I find this a hard estimation to make. Is this obvious? Also, how can I see surjectivity. Thanks for any help.
Hint: I assume your functions are $C^2$ on $[a,b]$. Observe the affine trick $$ u(\alpha)=\int_\alpha^\beta\left( \left(\frac{t-\beta}{\alpha-\beta} \right)u(t)\right)'dt. $$ Then differentiate the integrand and apply Cauchy-Schwarz to each summand. Then repeat the trick for $u(\beta)$, $u'(\alpha)$, and $u'(\beta)$.