Prove that $\frac{1}{\phi(D)}c=c\frac{1}{\phi(0)}$

Question

Could someone prove the following?

$$\frac{1}{\phi(D)}c=c\frac{1}{\phi(0)}$$

where $D$ is ${\frac{d}{dx}}$ and $c$ is a constant.

for example $$\frac{1}{D^4+2D+3}c=c\frac{1}{0+0+3}=\frac{c}{3}$$

Answer 1

As stated in the comments, you should specify what class of $\phi$ you are considering or what restrictions you impose. Anyway, proceeding a bit vaguely and assuming a "nice class" (for example, at least the inverse of $\phi(D)$ and $\phi(0)$ should be defined) you can transform your equation to (multiplying by $\phi(D)$ on the left and $\phi(0)$ on the right)

$$ c \phi(0) = \phi(D) c $$

now you can easily show that the above equation is true for, e.g. polynomials, i.e. when

$$ \phi(x) = \sum_{n=0}^N \alpha_n x^n $$