Here, the domain of $L_0$ is $\mathcal{C}_0^{\infty} (a,b)$ (test functions on $(a,b)$). Also, $L_0$ is an operator in the $L_2 (a,b)$ space i.e. $L_0 : \ \mathcal{C}_0^{\infty} (a,b) \subset L_2 (a,b) \rightarrow L_2 (a,b)$
I already proved it for myself that $\langle L_0 u, u \rangle$ must be real-valued
But how on earth should I show that $L_0$ is positive? I know that if I take $u=0$, then $\langle L_0 u, u \rangle = 0$. But how can I find out, whether $\langle L_0 u, u \rangle > 0$ if I take $u \neq 0$?
When $m$ is even, let $m=2n$ and it follows $$\langle L_0u,u\rangle=(-1)^n\langle u,u^{(2n)}\rangle=(-1)^{2n}\langle u^{(n)},u^{(n)}\rangle=\|u^{(n)}\|^2\geq0.$$ This equals zero iff $u^{(n)}\equiv0$, which holds iff $u\equiv0$ since $u\in\mathcal C_0^\infty(a,b)$.
If $m$ is odd, there are counterexamples: Consider $L_0=(-i)u'$ on $\mathcal C_0^\infty(-1,1)$. Let $u$ be a real-valued test function in $\mathcal C_0^\infty(-1,1)$, for example, $$u=e^{1/(x^2-1)}\ \text{if}\ x\in(-1,1),\quad u=0\ \text{otherwise},$$ then it follows $$\langle L_0u,u\rangle=(-i)\int uu'=-\frac i2\int\big(u^2\big)'=-\frac i2\big(u^2(1)-u^2(-1)\big)=0,$$ but $u\neq0$ on $(-1,1)$. In general, let $m=2n+1$ and we have $$\langle L_0u,u\rangle=(-i)\langle u^{(n)},u^{(n+1)}\rangle.$$ Since $\langle L_0u,u\rangle$ is real, $\langle u^{(n)},u^{(n+1)}\rangle$ is pure imaginary and therefore $0$ if $u$ is real-valued.