Show that there is a solution of the Laplace equation $(\mu-A)p=f$ Let $C_0(\mathbb R)$ denote the space of continuous functions vanishing at infinity equipped with the supremum norm, $B$ be a contractive linear operator on $C_0(\mathbb R)$ and $$Af:=\lambda(Bf-f)\;\;\;\text{for }f\in C_0(\mathbb R).$$


  Are we able to show that for some $\mu&gt;0$, there exists a dense subspace $D\subseteq C_0(\mathbb R)$ (maybe even $D=C_0(\mathbb R)$) such that for all $f\in D$, $$\underbrace{(\mu-A)}_{=:\:A_\mu}p=f\tag1$$ has a solution $p\in C_0(\mathbb R)$?


EDIT: Maybe it is as simple as follows: By the Lumer-Phillips theorem, $A$ is the generator of a strongly continuous contraction semigroup $(T(t))_{t\ge0}$ and maybe we can show that $$p:=\int_0^\infty e^{-\mu t}T(t)f\:{\rm d}t$$ is the desired solution of $(1)$ for a given $f$.
The equality $(\mu-A)p=f$ is equivalent to (assume $\lambda + \mu \neq 0$)

$$ p= \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}.$$

Let $\Phi : X\to X$, $\Phi(p) = \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}$. Then 

$$\| \Phi (p_1) - \Phi (p_2)\| = \frac{\lambda}{|\lambda + \mu|} \| B p_1 - B p_2\|&lt; \frac{\lambda c}{|\lambda + \mu|}\| p_1 - p_2\| $$
for some $c&lt;1$ since $B$ is contractive. 

Note that $\Phi$ is also contractive if 

$$ \frac{|\lambda|c}{|\lambda + \mu|} &lt; 1 \Leftrightarrow \mu &gt; c|\lambda | - \lambda \text{ or } \mu &lt; -c|\lambda| - \lambda$$

so for these choices of $\mu$, there is $p\in X$ so that $\Phi(p)= p$, which is the same as the equality you want.
The equality $(\mu-A)p=f$ is equivalent to (assume $\lambda + \mu \neq 0$)

$$ p= \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}.$$

Let $\Phi : X\to X$, $\Phi(p) = \frac{\lambda}{\lambda + \mu} Bp + \frac{f}{\mu+ \lambda}$. Then 

$$\| \Phi (p_1) - \Phi (p_2)\| = \frac{\lambda}{|\lambda + \mu|} \| B p_1 - B p_2\|&lt; \frac{\lambda c}{|\lambda + \mu|}\| p_1 - p_2\| $$
for some $c&lt;1$ since $B$ is contractive. 

Note that $\Phi$ is also contractive if 

$$ \frac{|\lambda|c}{|\lambda + \mu|} &lt; 1 \Leftrightarrow \mu &gt; c|\lambda | - \lambda \text{ or } \mu &lt; -c|\lambda| - \lambda$$

so for these choices of $\mu$, there is $p\in X$ so that $\Phi(p)= p$, which is the same as the equality you want.