How can $\sin(n \pi x)$ be a complete ONB of $L^2([0,1])$ when $-\Delta$ is positive on them As done in this question one can prove that $\sin(n\pi \cdot), n \ge 1$ is a orthonormal basis of $L^2([0,1])$.
We also have that $(-\Delta)( \sin(n\pi\cdot) )= (n\pi)^2 \sin(n\pi \cdot)$, which means that we can diagonalize $-\Delta$ w.r.t. $\sin(n \pi x)$ and have only positive eigenvalues.
From this I now conclude that $-\Delta$ is strictly positive on $L^2([0,1])$. This is a contradiction to the constants and the linear functions being in the kernel of $-\Delta$ and also in $L^2([0,1])$
 A: The operator domain comes with endpoint conditions $f(0)=f(1)=0$. That eliminates constants from being in the domain of the operator, unless those constants are $0$. If you choose endpoint conditions $f'(0)=f'(1)=0$, then you end up with a new basis of eigenfunctions $\{1,\cos(\pi x),\cos(2\pi x),\cdots\}$.
A: This would be a contradiction if $-\Delta$ were a bounded operator, but it's not (it's not even defined on all of $L^2([0,1])$.).
Write out the proof that it's a contradiction in detail, and you'll see there's a point where you use the non-fact that the operator is continuous.
A: 
From this I now conclude that $-\Delta$ is strictly positive on $L^2([0,1])$.

This is not true, because the representation of some element with respect to the mentioned ONB and the operator $-\Delta$ do not commute, because $-\Delta$ is not continuous operator!
As an example, let $f= 1_{[0,1]}$ be the indicator function of $[0,1]$, then
\begin{equation}
\int_0^1 \sin(\pi nx) \, \mathrm{d} x = \frac{1-(-1)^n}{\pi n}.
\end{equation}
Thus
$$1_{[0,1]}(x) = \sum_{n=1}^\infty \frac{1-(-1)^n}{\pi n} \sin(\pi nx)$$
in $L^2$-sense. Now, you cannot apply the operator 'coordinatewise'! The resulting sum is not convergent.
