I am reading the paper Uniqueness of the initial-value problem $u_t - \Delta \varphi u = 0$ by Brezis and Crandall.
At the beginning of the analysis, they introduce an operator $B g$ by $$ B g := (\epsilon - \Delta)^{-1} g$$ i.e., $B g$ is the unique solution of $$ \epsilon v - \Delta v = g \text{ in the sense of distributions}.$$
After introducing this operator, the authors claim that $B$ is "obviously symmetric".
I don't see this.
Attempt
I tried the following formal relations \begin{align} \langle f, Bg \rangle &= \epsilon^{-1} \langle f, \epsilon Bg \rangle \\ &= \epsilon^{-1} \langle f, g + \Delta Bg \rangle \\ \langle Bf, g \rangle &= \epsilon^{-1} \langle g, f + \Delta Bf \rangle \end{align}
Therefore
$$\langle f, Bg \rangle - \langle Bf, g \rangle =\epsilon^{-1} [\langle f, \Delta Bg \rangle - \langle g, \Delta Bf \rangle] $$
I don't see how to show that this difference equals zero.