Let $X$ be the completion of the space of smooth, compactly supported real-valued functions on $\mathbb R$ under the norm $$\|f\|_X^2=\int_{\mathbb R} \left(\frac{df}{dx}\right)^2 + f^2.$$ Let $Y=L^2(\mathbb R)$.
Prove the map $T:X \rightarrow Y$ given by $T(f)= \frac{df}{dx} - \frac{x}{\sqrt{1+x^2}}f $ has closed image and $1$-dimensional cokernel.
This is a qualifying exam problem. See Day 2, problem 6(d).
I noticed that $f\mapsto f'-f$ was an isometry, but I'm not sure this is useful. I don't see how to use the fact that the operator is a small perturbation of an isometry. I also tried to identify the adjoint, but I came to the equation $$(T^*g)''+T^*g=g'-\frac{x}{\sqrt{1+x^2}}g$$ after writing out the definition of adjoint and using integration by parts. This seems intractable.
I would appreciate any advice about how to proceed.
I think this may be a special case of some more general theorem about Sobolev spaces, but I am unfortunately ignorant of the relevant theory.
Edit: I suspect my equation for the adjoint is wrong. For if $T^*g=0$, then solving the resulting differential equation shows $g=Ce^{\sqrt{x^2+1}}$, which is of course not in $L^2$. So it seems the kernel of the adjoint is trivial, meaning the cokernel of $T$ is trivial. This contradicts the problem statement.
But its derivation seems correct. We must have $$\int (Tf)g=\int f'(T^*g)'+f(T^*g)=\int f((T^*g)'' + (T^*g))$$ for all compactly supported smooth $f$, so the equation follows, or so it seems. Something is clearly amiss.
The operator $$ Tf = f'-\frac{x}{\sqrt{1+x^{2}}}f $$ is a bounded operator from $X$ to $Y$ because $$ \begin{align} |Tf| & \le |f'|+|f|,\\ |Tf|^{2} & \le |f'|^{2}+|f|^{2}+2|f'||f| \\ & \le 2|f'|^{2}+2|f|^{2} \\ \|Tf\|_{Y}^{2} & \le 2\|f\|_{X}^{2}. \end{align} $$ To look at the range, let $g \in L^{2}(\mathbb{R})$ and try to solve for $f\in X$ such that $$ f'-\frac{x}{\sqrt{1+x^{2}}}f = g. $$ This requires an integrating factor: $$ \exp\left\{-\int \frac{x}{\sqrt{1+x^{2}}}dx\right\} = e^{-\sqrt{1+x^{2}}}. $$ The solution must satisfy $$ \{e^{-\sqrt{1+x^{2}}}f\}' = e^{-\sqrt{1+x^{2}}}g. $$ So the solution could be either of the following: $$ f(x) = e^{\sqrt{1+x^{2}}}\int_{-\infty}^{x}e^{-\sqrt{1+t^{2}}}g(t)dt \\ f(x) = -e^{\sqrt{1+x^{2}}}\int_{x}^{\infty}e^{-\sqrt{1+t^{2}}}g(t)dt. $$ Subtracting these two forms gives $$ 0 = \int_{-\infty}^{\infty}e^{-\sqrt{1+t^{2}}}g(t)dt. $$ The above condition becomes necessary in order to have $f \in X$. Conversely, if the above condition is met, then either of the two forms of solutions are the same, from which you can deduce that $f \in X$ as required. So the co-dimension of the range is $1$, which means the range is closed. It's easy to check that $T$ has trivial null space because $f'-\frac{x}{\sqrt{1+x^{2}}}f=0$ implies $f=Ce^{\sqrt{1+x^{2}}}$ which is not in $X$ unless $C=0$. So $\mathcal{N}(T)=\{0\}$.