Does an Operator on $C^2[0,1]$ exist?

Question

Assume you have a linear operator $A:C^2[0,1]\rightarrow C^2[0,1]$ s.t.

$(1-A)f(x)=c$ if $f(x)=c$ a constant function, which means $A\mathbb{1}=0$ and $(1-A)\tilde{f}(x)=const.$ (a constant function) if $\tilde{f}$ is not constant. Can we show that such an operator doesn't exist?

Answer 1

The operator $$Af=f-f(0)$$ satisfies the required assumptions. More generally for a probability measure $\mu $ on $[0,1]$ the operator $$Af=f-\int\limits_0^1f(t)\,d\mu(t)$$ satisfies the assumptions.