Is this operator symmetric and, if so, how to manifest the reality of its $L^2$ weighted norm?

I am working with an integral within the context of a Carleman estimate, and am trying to manifest its reality (with the later goal of finding a lower bound for $$-S$$ in the $$L^2$$ sense) but am having trouble. Although I believe the operator is symmetric from my calculations, there might be small errors, so I wanted to first ask the question of symmetricity of the operator $$S$$ that I will implicitly define below. If the answer to the question is yes, I am interested in how to manifest the reality of its corresponding $$L^2$$ weighted norm. For context, let $$f\in C_0^\infty(\mathbb{R}\times[0,1])$$ take values in $$\mathbb{C}$$, $$\alpha\in\mathbb{R}$$, $$\phi\in C^\infty([0,1])$$, take values in $$\mathbb{R}$$ and define the function $$\psi(x,t)=x+\phi(t)$$. I am interested in the following integral $$\begin{split} \int f^\dagger Sf &:= \alpha \int 192\alpha^4 \psi^3 f^\dagger \partial_x f + \alpha^248\psi^4 f^\dagger \partial_ {xx}f+6i\alpha^2 \psi^2 \phi'(t)f^\dagger \partial_x f - 48\alpha^2 \ \psi f^\dagger \partial_{xxx}f\\ &\quad-12 \alpha^2 \psi^2 f^\dagger \partial_{xxxx} f -12 \alpha^2 f^\dagger \partial_{xx}f+(1/2)i\phi'(t)f^\dagger \partial_{xxx}f + f^\dagger \partial^6_{x} f \\ &\quad+ \frac{1}{4}f^\dagger f\left(48\alpha^2 \psi^2 - 256 \alpha^6 \psi^6 +24i \alpha^2 \psi \phi'(t) -\frac{1}{4}\psi\phi'' - \frac{1}{4} \phi \right), \end{split}$$ where the integrals are computed over $$[0,1]\times \mathbb{R}$$, and the implicitly defined operator $$S$$ should be symmetric according to my calculations. My questions are:

1. Is the operator $$S$$ defined above indeed symmetric?

2. If yes, how to manifest the reality of the integral? In other words, how to show that the terms containing the imaginary unit $$i$$ either, e.g., disappear or turn into the real/imaginary part of some expression.

Edit 1: I added a pre-factor of $$\alpha$$ inside the integral

Edit 2: I added a forgotten factor of $$\alpha^2$$ to the term $$48\psi^4 f^\dagger \partial_ {xx}f$$

• I'm not sure why close votes were added? This question came up in the context of my research, and I believe it is on a similar level as some of the other well-received questions I've asked on this site. Commented Dec 16, 2021 at 23:48
• You are saying the operator S should be symmetric "according to your calculations," but you are not sure. This sounds like you are asking people to check your algebra. That is not research level, regardless of context, in my opinion, and this is why I voted to close. Commented Dec 18, 2021 at 2:50
• as I show in the answer box, this integral is complex in general, however it is real for $\alpha=1$ and $\phi(t)\equiv 0$; this may help you track down the error in your calculations. Commented Dec 19, 2021 at 13:24
• edit 2 is still complex, a suggestion for edit 3 to make it real is to replace $192\alpha^4$ by $192\alpha^2$. Commented Dec 19, 2021 at 23:52
• That makes sense. You could also replace $48\alpha^2$ by $48\alpha^4$ in the second term to make it real right? Commented Dec 20, 2021 at 0:17

The small print refers to the original expression of the OP, without the subsequent edits.

Let me try something simple: $$\alpha=0$$, $$\phi(t)\equiv 0$$, then $$I=\int_{-\infty}^\infty dx\int_0^1 dt\, f^\dagger Sf =\int_{-\infty}^\infty dx\int_0^1 dt\,\left(48x^4f^\dagger\frac{\partial^2 f}{\partial x^2}+f^\dagger\frac{\partial^6 f}{\partial x^6}\right),$$ and the question is whether this integral is real for a complex valued function $$f(x,t)$$.

The answer is no, for example, take $$f(x,t)=e^{-x^2}(1+ix^2)$$, then $$I=-\left(\frac{87}{4}+72 \,i\right) \sqrt{\frac{\pi }{2}}.$$

Let me now consider the revised integral of the OP,

$$\begin{split} I=\int f^\dagger Sf &:= \alpha \int 192\alpha^4 \psi^3 f^\dagger \partial_x f + \alpha^2 48\psi^4 f^\dagger \partial_ {xx}f+6i\alpha^2 \psi^2 \phi'(t)f^\dagger \partial_x f - 48\alpha^2 \ \psi f^\dagger \partial_{xxx}f\\ &\quad-12 \alpha^2 \psi^2 f^\dagger \partial_{xxxx} f -12 \alpha^2 f^\dagger \partial_{xx}f+(1/2)i\phi'(t)f^\dagger \partial_{xxx}f + f^\dagger \partial^6_{x} f \\ &\quad+ \frac{1}{4}f^\dagger f\left(48\alpha^2 \psi^2 - 256 \alpha^6 \psi^6 +24i \alpha^2 \psi \phi'(t) -\frac{1}{4}\psi\phi'' - \frac{1}{4} \phi \right), \end{split}$$

To check whether this is real I consider order by order in $$\alpha$$, $$I=\alpha I_1 + \alpha^3 I_3+\alpha^5 I_5+\alpha^7 I_7.$$ Each $$\alpha$$-independent term $$I_p$$ should be separately real.

$$\bullet$$ Start with $$I_1$$, $$I_1=\int_{-\infty}^\infty dx\int_0^1 dt\,\left(\tfrac{1}{2}i\phi'f^\dagger \frac{\partial^3 f}{\partial x^3} + f^\dagger \frac{\partial^6 f}{\partial x^6} -\tfrac{1}{16}(x+\phi)\phi''f^\dagger f - \tfrac{1}{16} \phi f^\dagger f\right).$$ Perform a partial integration with respect to $$x$$, keeping in mind that $$\phi$$ depends on $$t$$ only, $$I_1=\int_{-\infty}^\infty dx\int_0^1 dt\,\left(-\tfrac{1}{2}i\phi'f \frac{\partial^3 f^\dagger}{\partial x^3} + f \frac{\partial^6 f^\dagger}{\partial x^6} -\tfrac{1}{16}(x+\phi)\phi''f^\dagger f - \tfrac{1}{16} \phi f^\dagger f\right).$$ The second expression equals the complex conjugate of the first, hence $$I_1$$ is real.

$$\bullet$$ Continue with $$I_3$$, $$I_3=\int_{-\infty}^\infty dx\int_0^1 dt\,\left(12\left\{4(x+\phi)^4 f^\dagger \frac{\partial^2 f}{\partial x^2} - 4 \ (x+\phi) f^\dagger \frac{\partial^3 f}{\partial x^3}- (x+\phi)^2 f^\dagger \frac{\partial^4 f}{\partial x^4}\right\} +6\phi'\left[i (x+\phi)^2 f^\dagger \frac{\partial f}{\partial x}+ i(x+\phi) f^\dagger f\right]-12 f^\dagger \frac{\partial^2 f}{\partial x^2}+12 (x+\phi)^2f^\dagger f \right).$$ If you now take the complex conjugate and do partial integrations with respect to $$x$$, you see that the last two terms are separately real, the sum of the two terms between square brackets is real, but the sum of the three terms between curly brackets is not real.

$$\bullet$$ Next is $$I_5$$, $$I_5=192\int_{-\infty}^\infty dx\int_0^1 dt\, (x+\phi)^3 f^\dagger \frac{\partial f}{\partial x}.$$ Its complex conjugate is $$\bar{I}_5=-192\int_{-\infty}^\infty dx\int_0^1 dt\, \left[(x+\phi)^3 f^\dagger \frac{\partial f}{\partial x}+3(x+\phi)^2f^\dagger f\right].$$ This will in general be different from $$I_5$$, so it is complex. The remaining term $$I_7$$ has a real integrand $$\propto (x+\phi)^6 f^\dagger f$$, so it is real.

Conclusion: the corrected expression in the OP is still not real. One way to fix this is to add $$I_5$$ to $$I_3$$, so if instead of $$192\alpha^4$$ one would write $$192\alpha^2$$ in the very first term of the integral.

• Hi, I forgot to include that there is a pre-factor of $\alpha$ in front of the integrand. I thought it would be ok to neglect it since $\alpha$ is real, but I didn't think of what would happen if $\alpha$ was zero. Sorry about that, I edited my post to add that. Commented Dec 18, 2021 at 16:11
• this prefactor of $\alpha$ does not change anything: just calculate the integral to first order in $\alpha$, and you find for the example I considered that $$I=-\alpha\left(\frac{87}{4}+72 \,i\right) \sqrt{\frac{\pi }{2}}+{\cal O}(\alpha^2),$$ and the terms of order $\alpha^2$ cannot cancel the imaginary part of order $\alpha$. Commented Dec 18, 2021 at 17:24
• Hi. Thank you for your efforts in solving this problem. I'm a bit embarrassed to admit it but I forgot a pre-factor of $\alpha^2$ in one of the terms in the integrand (it's now fixed), which explains why with $\alpha=1$, it would be symmetric. However, there is still the issue of the $\mathcal{O}(\alpha^0)$ terms in the integrand for $\phi(t)$ not identically zero. Commented Dec 19, 2021 at 15:23
• OK, so I worked on your latest corrected expression, which is still not real -- the term of order $\alpha^5$ is complex. Commented Dec 19, 2021 at 18:29
• Thank you for your efforts, I'll try to see where my error is. Commented Dec 19, 2021 at 18:32