How to justify solving $f(x+1) + f(x) = g(x)$ using this spectral-like method?

Question

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.

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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 $f(x+1) + f(x) = (x+2)^2$ !

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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.

Answer 1

First, if $g(x)$ is an order $n$ polynomial, we can find a unique order $n$ polynomial $f(x)$ that solves your functional equation by equating the polynomial coefficients on both sides of the equation, but on the other hand your solution method is very clever and interesting. For the remainder, consider $f$ and $g$ to be Fourier transformable distributions (test functions that fall off faster than $1/x^n$ should work).

$T$ is a unitary operator with $(T^\dagger f)(x)=f(x-1)$ and eigenvector $e^{ikx}$ has eigenvalue $e^{ik}$, which equals $-1$ for $k=(2n+1)\pi$. Now suppose $g(x)=e^{ikx}$ with $k \neq (2n+1)\pi$, then your series for $(I+\lambda T)^{-1} g$ becomes $(1-\lambda e^{ik}+\lambda^2 e^{i2k}-...)g=(1+\lambda e^{ik})^{-1}g$. Although the series does not converge for $\lambda = 1$ the limit exists. For $k=(2n+1)\pi$ the series becomes $(1+\lambda +\lambda ^2+\lambda ^3+...)g=(1-\lambda )^{-1})g$, which does not have a good limit as $\lambda \mapsto 1$. In fact, if $(I+T)g=0$, then $(I+T)^2f=0$, but for unitary $T$ $(I+T)^2f=0$ does not have any solutions that do not already solve $(I+T)f=0$. The moral of the story is that your solution method fails if the Fourier transform of $g(x)$ is non-zero for any $k=(2n+1)\pi$.

Finally, as to why your series works for polynomials, the Fourier transform of $x^n$ is $(i\frac{d}{dk})^n\delta (k)$, which does not contain any forbidden frequencies.