Let $L_0 = (-i)^m \frac{d^m u}{dx^m}$. How can I prove, that $L_0$ is a non-negative operator?

Question

1. The minimal operator $L_0 = (-i)^m \frac{d^m u}{dx^m}, \ u \in \mathcal{C}_0^{\infty} (a,b)$ for $- \infty \le a < b < \infty$ induced in $L_2 (a,b)$ by the differential form $l[u] = (-i)^m \frac{d^m u}{dx^m}$ is non-negative

2. A self-adjoint extension $A$ of $L_0$ is given by $$Au = (-i)^m \frac{d^{2n} u}{dx^{2n}}, \ u \in dom A$$ where $dom A$ consists of all $u \in W_2^m (a,b)$ such, that

$$f(a) = f' (a) = ... = f^{m-1} (a) = 0 \ if \ a > - \infty$$

and

$$f(b) = f' (b) = ... = f^{m-1} (b) = 0 \ if \ b < \infty$$

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My attempt

1. We want to prove, that $\langle L_0 u, u \rangle \ge 0$. We know, that if it has to be non-negative, then it has to be real, thus $\langle L_0 u, u \rangle = \overline{\langle L_0 u, u \rangle} = \langle L_0^* u, u \rangle$. This again means, that $L_0 = L_0^*$.

$$\langle L_0 u, u \rangle = (-i)^m \int_a^b \frac{d^m u}{dx^m} \overline{u} dx = (-i)^m (\frac{d^{m-1}u}{dx^{m-1}} \overline{u} \ |_a^b - \int_a^b \frac{d^{m-1}u}{dx^{m-1}} \overline{u'}dx) = ... $$ Because the support of $u$ here is $(a,b)$, this means that ultimately, we get: $$... = (-i)^m (-1)^m \int_a^b u \overline{\frac{d^m u}{dx^m}}dx = i^m \int_a^b u \overline{\frac{d^m u}{dx^m}}dx = \int_a^b u \ \cdot \ \overline{(-i)^m \frac{d^m u}{dx^m}}dx = \langle u, L_0 u \rangle$$ I proved that indeed $\langle L_0 u, u \rangle$ is real-valued. But how do I prove that it is non-negative, too?

2. My idea would be to use integration by parts, too. But the problem here is that it's the $2n$-th derivative, and not the $m$-th derivative. So that's where I'm stuck

Answer 1

I am pretty sure it is a misprint. Is there any way you can remove the bounty? 350 points for a misprint seem like a big waste.

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Anyway, in case it is not a misprint, let's prove that $2n=m$. Since $A$ is an extension of $L_0$, the two operators coincide on $C_0^\infty(a,b)$. Hence, $$\langle Au-L_0u,v\rangle-0$$ for all $u,v\in C_0^\infty(a,b)$. This implies that $$\int_a^b\left(\frac{d^mu}{dx^m}(x)-\frac{d^{2n}u}{dx^{2n}}(x)\right)\bar v(x)\,dx=0$$ for all $u,v\in C_0^\infty(a,b)$. Taking $v=\frac{d^mu}{dx^m}-\frac{d^{2n}u}{dx^{2n}}$, you have $$\int_a^b\left|\frac{d^mu}{dx^m}(x)-\frac{d^{2n}u}{dx^{2n}}(x)\right|^2\,dx=0.$$ Hence, $\frac{d^mu}{dx^m}(x)-\frac{d^{2n}u}{dx^{2n}}(x)$ for every $x$ and every $u$, which clearly implies that $2n=m$. Now you can continue as in your comment above.