Let's say that I want to find solutions $f\in C(\Bbb R)$ to the equation
$$ f(x+1) + f(x) = g(x) $$
for some $g\in C(\Bbb R)$. I can write $f(x+1) = (Tf)(x)$ where $T$ is the right shift operator and rewrite the equation suggestively as $$ (I+ T)f=g. $$ Formally, I can say that the solution of this equation is
$$ f= (I+ T)^{-1} g. $$
Of course, I am aware that there are infinitely many solutions to the equation but please bear with me for a moment here.
By the theory of operator algebra, if $f,g$ are from some nice Banach space $X$ and our linear operator $T:X\to X$ satisfies $\|T\|<1$, then we have $$ f = \left(\sum_{n=0}^\infty (-T)^n \right)g. $$
However, it is not unreasonable to expect that we should have $\|T\|=1$ for a right-shift operator in most reasonable function spaces so let's try to solve the equation $$ f= (I+\lambda T)^{-1} g. $$ for $\lambda <1$ first then we'll take $\lambda\to 1$. Note that all the steps until now is purely formal since $C(\Bbb R)$ is not a normed space.
For a concrete example, let's say we take $g(x) = (x+2)^2$. The previous method says that we first calculate (for $\lambda<1$) $$\begin{align} f(x) &= \left(I - \lambda T + \lambda^2 T^2 - \dots \right) g(x) \\ &= (x+2)^2 -\lambda (x+3)^1 + \lambda^2 (x+4)^2 + \dots \\ &= \left(1-\lambda+\lambda^2-\dots \right)x^2 + \left(2-3\lambda+4\lambda^2-\dots \right)2x + \left(2^2-3^2\lambda+4^2\lambda^2-\dots \right) \\ &= \frac{1}{1+\lambda} x^2 + 2 \frac{2+\lambda}{(1+\lambda)^2} x + \frac{4+3\lambda + \lambda^2}{(1+\lambda)^3}. \end{align}$$ We shall be brave here and substitute $\lambda=1$ even though the series doesn't converge there. This gives $$ f(x) = \frac 12 x^2 + \frac 32 x + 1 $$ but voilà, for some mysterious reasons unknown to me, this $f$ actually solves our original equation $f(x+1) + f(x) = (x+2)^2$ !
My question is simply:
What are the hidden theories behind the miracle we observe here? How can we justify all these seemingly unjustifiable steps?
I can't give you a reference to this method because I just conjured it up, thinking that it wouldn't work. To my greatest surprise, the answer actually makes sense. I am sure that similar method is probably practiced somewhere, probably by physicists.
Some points worth mentioning:
1.) $C(\Bbb R)$ is probably not the right space to work with since it's not normed. However, I want my answer to be a continuous function on $\Bbb R$ so some form of continuity assumption is needed for our space $X$.
2.) Norming $C(\Bbb R)$ with $L^\infty$ norm is not the way to go since it is possible that $f$ is unbounded, as our example shows.
3.) The solution $f$ is not unique since the kernel of $(I+T)$ consists of all $C(\Bbb R)$ functions $h$ such that $h(x+1)=-h(x)$, e.g. $\sin(\pi x)$.
4.) All the series expansions for $\lambda$ in my example converges when $|\lambda| <1$ but not at $\lambda=1$ but for some reason the result checks out.