I have to show that the operator
$$A:\mathcal{C}([0,1],\mathbb{C})\rightarrow\mathcal{C}([0,1],\mathbb{C}),$$
$$f\mapsto f'',$$
with domain
$$D(A)=\left\{f\in\mathcal{C}^2([0,1])|f'(0)+\alpha f(0)=f'(1)+\beta f(1)=0\right\}$$
is dissipative, i.e.
$$||\lambda f-f''||\geq \lambda ||f||\;\;\forall \lambda >0,$$
where $||f||=||f||_{\infty}=\sup_{[0,1]}|f|=\max_{[0,1]}|f|$.
Since all functions considere are continuous I know they attain their maximum in $[0,1]$, say at $s$. When $s\in(0,1)$ I managed to show dissipativity by using the equivalent characterization:
An operator $T:D(T)\subseteq X\rightarrow X$ is dissipative iff
$$\exists x^*\in \mathcal{I}(x)|\Re(x^*(Tx))\leq 0,$$
where $\mathcal{I}(x)=\{x^*\in X^*|x^*(x)=||x||^2\}$.
Indeed, if we denote by $\delta_s$ the Dirac distribution then $f^*=\bar{f}(s)\delta_s$ does the trick. However, if $s=0$ (I suspect the other case is analogous), I can't seem to find the appropriate $f^*$. My strategy above was inspired by the elementary result from calculus that the second derivative of a real-valued function is non-positive at a local critical maxima, but this is of course false in the case of a maxima attained at the boundary such in my case.