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 $48\psi^4 f^\dagger \partial\_ {xx}f$



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


